01 Stoichiometry Premium VersionBy Bomvirtual.com / June 8, 2025 0% 0 votes, 0 avg 88 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980 Created by Bomvirtual.com 01 Stoichiometry Premium Version 1 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 1) The molecular weight of a gas is 45. Its density at STP is A) 2 B) 5.7 C) 22.4 D) 11.2 The density of gas = molecular wt. of metal / volume = 45/22.4 = 2gm/L 2 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 2) 1 mol of CH₄ contains A) 3.0 g of carbon B) 4 g atom of Hydrogen C) 6.02 × 10²³ atoms of H D) 1.81×10²³ molecules of CH₄ 1 mol of CH4 contains 4 g atom of Hydrogen 3 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 3) How many oxygen atoms are present in 278g of Hydrated Ferrous Sulphate? (FeSO₄.7H₂O = 278 amu) ETEA 2016 A) 6.023 × 10²³ B) 6.525 × 10²⁴ C) 2.408 × 10²³ D) 6.023 × 10²² Molar mass of FeSO₄·7H₂O: = 56 + 32 + 64 + 126 = 278 g/mol Since the given mass is 278g, the number of moles is: 278 𝑔/ 278 =1 mole, FeSO₄·7H₂O: have 11 atoms. , Oxygen atoms: 11 x NA = 6.525 ´ 1024 4 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 4) The amount of products that is actually produced during a chemical reaction by performing an experiment is called _____ ETEA – 2022 A) Actual yield B) Theoretical yield C) Mole D) Percent yield The term for the amount of product actually obtained from a chemical reaction is the Actual yield. It is often compared to the theoretical yield to calculate the percent yield of a reaction. 5 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 5) Which of the following has least mass? A) 3×10²³ atoms of C B) 1 mole of S C) 2 g atom of nitrogen D) 7.0 g of Ag 2gm atom of nitrogen = 28gm , 6 x1023 atoms of C has mass = 12gm, 3 x1023 atoms of C has mass = 6 gm, 1mole of S has mass = 32gm, 7.0gm of Ag So, lowest mass = 6gm of C. 6 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 6) Molecular mass of the compound is 60 and its empirical formula is CH₂O. What will be the molecular formula of the compound? ETEA – 2022 A) C₂H₆O₂ B) C₂H₈O₂ C) C₆H₁₂O₆ D) C₂H₄O₂ Given the empirical formula CH2O and a molecular mass of 60. The molecular formula is a multiple of the empirical formula. The molar mass of the empirical formula is calculated as C = 12.01, H2 = 2.02, O = 16.00, so total = 30.03 g/mol. The ratio of the molecular mass to the empirical formula mass will give us the multiplier to find the molecular formula. 7 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Limiting and excess reagent 7) The reactant which consumes first in a chemical reaction is called? ETEA – 2023 A) Consumed reactant B) Non-reacted reactant C) Excess reactant D) Limiting reactant The reactant which consumes first in a chemical reaction is called Limiting reactant 8 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 8) In a vessel, 10g N₂, 10g H₂, and 10g O₂ are present. Which one will have the least number of atoms? ETEA – 2020 A) O₂ B) Both H₂ & N₂ C) N₂ D) H₂ Moles of N₂ = 10g / 28g/mol = 0.357, Moles of H₂ = 10g / 2g/mol = 5, Moles of O₂ = 10g / 32g/mol = 0.3125, Since each molecule contains two atoms, the total number of atoms is:, N₂ atoms = 0.357 × 2 = 0.714, H₂ atoms = 5 × 2 = 10, O₂ atoms = 0.3125 × 2 = 0.625, The gas with the least number of atoms is O₂. 9 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 9) What is the mass of 1 mole of calcium carbonate (CaCO₃)? KP-MDCAT 2024 A) 50g B) 100g C) 125g D) 75g Molar mass of CaCO₃ = 40 + 12 + (3×16) = 100 g/mol 10 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 10) The mass of a molecule of water is A) 3 × 10⁻²⁵ kg B) 1.5 × 10⁻²⁶ kg C) 2.5 × 10⁻²⁶ kg D) 3 × 10⁻²⁶ kg 6 x1023 molecules has mass = 18gm, 1 molecules has mass 18/6 x1023 = 3 x 10-23 gm 11 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 11) Which one of the following pairs of gases contains the same number of molecules? A) 8 g of O₂ and 22 g of CO₂ B) 28 g of N₂ and 22 g of CO₂ C) 32 g of O₂ and 32 g of N₂ D) 16 g of O₂ and 14 g of N₂ 16 g of O2 = 16/32 = 0.5 mol, 14 g of N2 = 14/28 = 0.5 mol 12 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 12) 19.7 kg of gold was recovered from a smuggler. How many atoms of gold were recovered (Au=197)? A) 6.023×10²⁴ B) 6.023×10²³ C) 6.023×10²⁵ D) 100 Amount of gold = 19.7kg = 19.7 x1000gm =19700gm, No. of moles = 19700/197= 100, No. of atoms =100 x 6.023×1023 , = 6.023 x 1025 atoms. 13 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 13) 117.0g of NaCl have A) 1.204 × 10²⁴ formula units of NaCl B) 12.04 × 10²² formula units of NaCl C) 1.204 × 10²³ molecules of NaCl D) 6.023 × 10²³ molecules of NaCl n = 117/58 = 2 mol, 2.0 moles contain:, 2.0× ( 6.022 × 1023 ) = 1.204 × 10 2 4 formula units 14 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 14) Cylinder “A” contains 4.6 grams of C₂H₅OH and cylinder “B” has 3 grams C₂H₆: ETEA 2016 A) Both cylinders have the equal number of hydrogen atoms B) Both cylinder A and B have equal number of molecules C) Cylinder A has greater number of molecules than cylinder B D) none of these Cylinder A (C₂H₅OH, Ethanol) Molar mass of C₂H₅OH = 46 g/mol, Moles in A = 4.6 g / 46 g/mol = 0.1 moles, Cylinder B (C₂H₆, Ethane), Molar mass of C₂H₆ = 30 g/mol, Moles in B = 3 g / 30 g/mol = 0.1 moles, Since both cylinders contain 0.1 moles, and 1 mole contains Avogadro’s number of molecules, both A and B have an equal number of molecules. 15 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 15) How many mole of helium gas occupy 22.4 L at 0°C at 1 atm. pressure? A) 1 B) 0.9 C) 1.11 D) 0.11 1mole of any gas at STP occupies 22.4L. 16 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 16) One mole of which of the following bucky ball will have more molecules? ETEA – 2019 A) C₆₀ B) all same C) C₅₀ D) C₂₀ One mole of any substance, including buckyballs, contains Avogadro’s number of molecules, so they all have the same number of molecules 17 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 17) The number of molecules of CO₂ present in 44g of CO₂ is A) 3.023×10²³ B) 12.023×10²³ C) 6.023×10²³ D) 3.023×10¹⁰ 1 mol CO2 = 44g = 6.023 x10 23 molecules 18 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 18) 1 amu is equal to 1.661 × 10⁻²⁴g, then 1.0 g will be equal to: ETEA 2012 A) 6.022 × 10⁻²⁴amu B) 6.022 × 10²³amu C) 6.022 × 10⁻²³amu D) 6.022 × 10²⁴amu number of amu = 1/1.61 x 10-24 = 2.023×1023 19 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 19) When 4 g of magnesium was heated in excess of oxygen. Calculate the theoretical yield of magnesium oxide (MgO). KP-MDCAT 2024 A) 3.7g B) 5.4g C) 6.6g D) 4.2g 2Mg + O₂ → 2MgO; 4g Mg = 0.165 mol → 0.165 mol MgO = 6.6g 20 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 20) What are the Avogadro’s number of particles in 0.25 moles of CO₂? NUMS 2020 A) 1.505 × 10²³ B) 2.00 × 10²³ C) 1.505 × 10¹⁵ D) 6.022 × 10⁷ 1 mol = NA → 0.25 mol = NA /0.25 = 1.505 × 1023 21 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 21) 1 mole of any substance contains ____ substances NUMS 2020 A) 6.02 × 10²⁴ B) 6.02 × 10²³ C) 3.01 × 10²³ D) 6.02 × 10²² 1 mole of any substance contains 6.02 × 1023 22 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 22) The volume occupied by 4.4 g of CO₂ at STP is A) 0.224 L B) 2.24 L C) 22.4 L D) 0.1 L 44g CO2 occupies 22.4L at STP, 4.4g CO2 occupies 22.4×4.4/44 = 2.24 L 23 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 23) 60 a.m.u of C-12 contain carbon: atoms ETEA 2016 A) 60 × 6.02 × 10²³ B) 5 × 6.02 × 10²³ C) 5 D) 60 number of atoms = 60/12 = 5 atom s 24 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 24) Efficiency of chemical reaction can be checked by calculating KP-MDCAT 2024 A) Theoretical yield B) Actual yield C) Percentage yield D) Amount of the reactant unused Percentage yield compares actual yield to theoretical yield, showing reaction efficiency 25 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 25) The numbers of moles of BaCO₃ which contain 1.5 moles of oxygen atoms is A) 3 B) 0.5 C) 1 D) 6.023×10²³ 3 moles of oxygen is that in 1 mole of BaCO3, 1.5 moles of oxygen is that in mole of BaCO3 = 1.5/3 = 0.5 26 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 26) The volume occupied by 3.2g of oxygen gas at S.T.P is: ETEA – 2023 A) 22.4 dm³ B) 2.24 dm³ C) 1.12 dm³ D) 24 dm³ 32 g = 22.8 dm3, .2 g = 2.28dm3 27 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 27) Which of the following contain the same number of molecules as 22 gram of carbon dioxide? ETEA – 2022 A) 32g of oxygen gas B) 2g of hydrogen gas C) 71g of chlorine gas D) 9g of water 22/44 = 0.5 moles CO2., 9/18 = 0.5 g water 28 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 28) The number of gram atoms in 3g Hydrogen atoms is the same as the number of gram atoms in 48g of ETEA 2018 A) N B) O C) C D) O₂ Moles = 3g / 1g/mol = 3 gram atoms Now, we check which element has 3 gram atoms in 48g: Nitrogen (N): 48g / 14g = 3.43 gram atoms, Carbon (C): 48g / 12g = 4 gram atoms, Oxygen (O): 48g / 16g = 3 gram atoms 29 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 29) How many grams of CO₂ can be produced by thermally decomposing 10 moles of ZnCO₃(S)? KP-MDCAT 2024 A) 320 B) 440 C) 400 D) 360 ZnCO₃ → ZnO + CO₂. 10 moles ZnCO₃ produces 10 moles CO₂ = 10 × 44g = 440g 30 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Limiting and excess reagent 30) Consider the following reaction involved in the manufacture of Urea: CO₂ + 2NH₃ → NH₂COONH₄ If 22.0g of CO₂ react with 34 g of ammonia to form ammonium carbamate, the reaction is taken as irreversible and go to completion. Identify the limiting reagent and the amount of carbamate formed: A) CO₂, 39g B) NH₃, 78g C) CO₂, 78g D) NH₃, 39g Moles of CO2=22/44=0.5 mol, Moles of NH3=34/17=2.0 mol, CO2 + 2NH3 NH2 COONH4, The reaction requires 1 mol of CO2 to react with 2 mol of NH3., For 0.5 mol CO2, we need 1.0mol of NH3, We have 2.0mol of NH3, which is more than enough. Thus, CO2 is the limiting reactant., 1 mol of CO2 produces 1mol of NH2COONH4., 00.5mol of CO2 produces 0.5mol of NH2COONH4., Molar mass of NH2COONH4 = 76g/mol., Mass formed = 0.5×76 =38 g 31 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 31) A ring contains 1.2 gram of diamond, the number of carbon atoms in the ring are: ETEA 2016 A) NA B) NA/2 C) 1.2NA D) NA/10 n = 1.2/12 = 0.1 mol →no of atoms = 0.1 x NA = NA/10 32 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 32) A gas at STP contains only 6.023×10²³ atoms and is monatomic it will occupy. ETEA 2010 A) 44.8L B) 1.2L C) 30.5L D) 22.4L 22.4dm3 = 22.4L = 6.023×1023 atoms 33 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 33) The efficiency of a chemical reaction can be expressed as: MDCAT 2020 A) Maximum yield B) % yield C) Theoretical yield D) Actual yield The efficiency of a chemical reaction is best expressed using percentage yield (% yield), which compares the actual yield (obtained in the experiment) to the theoretical yield (calculated from stoichiometry). % Yield = Actualyield/Theoreicayield×100 34 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 34) The sample of a compound contains 0.100g of hydrogen and 4.20g of nitrogen. The simplest formula for the compound is… ETEA-2005 A) NH₂ B) HN₂ C) NH₃ D) HN₃ To determine the simplest formula, we calculate the mole ratio of elements. Moles of H = 0.100g / 1g = 0.1 moles, Moles of N = 4.20g / 14g = 0.3 moles, Now, divide both values by the smallest number of moles (0.1):, H ratio = 0.1 / 0.1 = 1, N ratio = 0.3 / 0.1 = 3 So, the simplest formula is HN₃. 35 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 35) The number of molecules in 16 g of methane is A) 16/3×10²³ B) 16/6.023×10²³ C) 6.023×10²³ D) 3×10²³ 16 g = 1 mol = 6.023 x10 23 36 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 36) How many moles are there in 60g of sodium hydroxide (NaOH)? KP-MDCAT 2024 A) 4 B) 2 C) 8 D) 6 Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles = mass/molar mass = 60g/40g/mol = 1.5 moles (Note: This suggests there may be an error in the question or options as none match exactly) 37 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 37) Which of the following pairs have the same electronic structure? ETEA 2006 A) Ca & Ar B) Ag & Sn C) Ar & Cl D) Mg & Na Ar &Cl are Iso-electronic between both have 18 e. 38 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 38) Identify the correct option with same empirical formula for both compounds NUMS 2020 A) C₆H₁₂O₆ & CH₃COOH B) H₂O & H₂O₂ C) H₂S₂O₃ & H₂SO₄ D) C₆H₁₂ & C₆H₆ (a) H₂O & H₂O₂ → H₂O (empirical formula: H₂O), H₂O₂ (empirical formula: HO) → Different , (b) C₆H₁₂ & C₆H₆ → C₆H₁₂ (empirical formula: CH₂), C₆H₆ (empirical formula: CH) → Different, (c) H₂S₂O₃ & H₂SO₄ → H₂S₂O₃ (empirical formula: H₂S₂O₃), H₂SO₄ (empirical formula: H₂SO₄) → Different, (d) C₆H₁₂O₆ & CH₃COOH → C₆H₁₂O₆ (empirical formula: CH₂O), CH₃COOH (empirical formula: CH₂O) → Same 39 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 39) The volume of CO₂ produced by heating 33.5g Li₂CO₃ at room temperature and pressure is (Mr Li₂CO₃ = 67g/mol): ETEA 2018 A) 12.0 dm³ B) 22.4 dm³ C) 11.2 dm³ D) 24.0 dm³ 40 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 40) How many molecules are present in one gram of hydrogen? A) none of these B) 3.023×10²³ C) 2.69×10²³ D) 6.023×10²³ 2gm of hydrogen 6.023 x10 23 molecules, 1gm of hydrogen = 6.023 x10 23/2 =2.69 x10 23 41 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 41) Which compound with the given information has greater mass in Kg? ETEA 2018 A) 22.4 km³ N₂ at STP B) 6.02 × 10²³ molecules of C₂H₄ C) 2 mole of CO D) All have equal mass 22.4 km3 N2 at STP = 28 g = 2 mol CO 42 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 42) Na is Iso-electronic with: ETEA 2016 A) He B) Ne C) Mg D) Fe Na+ = 10 ȇ, Ne = 10e –––>Both are Iso-electronic 43 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 43) What is the number of hydrogen atoms in 5 moles of water? ETEA 2015 A) 6.023 × 10²³ B) 6.023 × 10²⁴ C) 5.0 × 10²³ D) 3.0115 × 10²⁴ # of Atoms = n × NA × # of Atoms in formula (Atomicity) = 5× 6.022 × 1023 × 2=6.023 × 1024 44 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 44) What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene? A) 96 kg B) 9.6 kg C) 2.8 kg D) 6.4 kg C2H4 + 2O2 → 2CO2 + 2H2O, 28gm C2H4 requires 64gm oxygen, 2.8 x103 gm C2H4 requires = 64/28 x 2.8×103 gm = 6.4 x1023 gm = 6.4kg. 45 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 45) Total number of atoms represented by the compound CuSO₄.5H₂O is A) 27 B) 5 C) 21 D) 8 CuSO4.5H2O = 1+1+9+10=21 46 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 46) Phosphorous exists in nature as a tetra atomic molecule. The number of atoms present in one gram molecule of phosphorous are: ETEA 2018 A) 3 × 6.023 × 10²³ B) None of the above C) 6.023 × 10²³ D) 2 × 6.023 × 10²³ 4 x NA = 4 x 6.023×1023 47 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Limiting and excess reagent 47) If 28.0g nitrogen gas is reacted with 8.0g of hydrogen gas to form Ammonia, the limiting reactant among the two will be: ETEA 2008 A) Both a & b B) N₂ C) H₂ D) None Moles of N2=28.0=1.0 mol, Moles of H2=8/2 =4.0mol, For 1.0 mol1.0mol of N2N2, we need 3.0 molof H2., We have 4.0mol of H2, which is more than enough. 48 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 48) Avogadro’s constant is the number of ETEA – 2019 A) Atoms in 1g of He B) electrons present in 2g H C) atoms in 24g of Mg D) molecules in 35.5g of chlorine Avogadro’s constant represents the number of entities in a mole of a substance, not specifically tied to any of the options given in the context of a specific weight or element. 49 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 49) The largest number of molecules is in A) 34g of water B) 54g of N₂O₅ C) 46g of CH₃OH D) 28g of CO₂ 50 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 50) Many elements have fractional atomic masses. This is because NUMS 2020 A) Atomic masses are average masses of isotopes B) Atomic masses are average masses of isotopes proportional to their abundance C) Atomic masses are average masses of isobars D) Mass of atom is itself fractional Fractional atomic masses arise because most elements exist as a mixture of isotopes, and the atomic mass listed in the periodic table is a weighted average based on their natural abundance. 51 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 51) Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85g of BaCO₃ (Atomic mass of Ba=137) will be A) 0.84 L B) 2.24 L C) 4.06 L D) 1.12 L BaCO3 → BaO +CO2, Molecular weight of BaCO3 = 137 +12 + 3x 16 =197, 197gm produces 22.4L at S.T.P., 9.85gm produces 22.4 x 9.85/197 = 1.12L at S.T.P. 52 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 52) Number of gm of oxygen in 32.2 g Na₂SO₄.10H₂O is A) 2.08 B) 22.4 C) 20.8 D) 2.24 Na2 SO4 .10H2O = 322 g molar mass., 322 g Na2 SO4 .10H2O = 224 g oxygen., 32.2 g Na2 SO4 .10H2O =22.4 g oxygen. 53 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Limiting and excess reagent 53) 20 cm³ CH₄ gas was burnt in 10 cm³ O₂ to produce CO₂ as: CH₄ + 2O₂ → CO₂ + 2H₂O The limiting reagent in this reaction is ETEA – 2019 A) None of the above B) CO₂ C) O₂ D) CH₄ Using stoichiometry, 1 mole of CH4 requires 2 moles of O2. Given the volumes (assuming equal conditions of temperature and pressure), O2 is the limiting reagent because 10 cm³ of O2 can react completely with 5 cm³ of CH4, but there’s 20 cm³ of CH4 available 54 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 54) The weight of 1×10²² molecules of CuSO₄.5H₂O is A) 4.159 g B) 41.59 g C) None of these D) 415.9 g Molecular weight of CuSO4.5H2O = 63.5+32+64+90=249.5, 6 x1023 molecules has weight = 249.5gm, 1 x1022 molecules has weight = 249 x1 x1022 /6 x1023 = 4.158 55 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 55) Theoretical yield is always: ETEA 2015 A) Both are equal B) None of the above C) Less than practical yield D) Greater than actual yield The theoretical yield is the maximum possible amount of product that can be obtained from a chemical reaction, based on stoichiometric calculations. However, in real experiments, the actual yield is usually less due to factors like incomplete reactions, side reactions, or losses during processing. 56 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 56) The number of moles of CO₂ which contain 8 gm of oxygen are NUMS 2020 A) 1 B) 4.5 C) 0.25 D) 0.5 The molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol, Oxygen contributes 32g (2 × 16g) per mole of CO₂, 1 mol CO2 have 32 g Oxygen so 0.25 mol CO2 will have 8 g oxygen 57 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 57) Choose the correct relation about the percent yield. It is equal to: ETEA 2014 A) (Actual yield / Theoretical yield) × 10⁶ B) (Theoretical yield / Actual yield) × 100 C) (Actual yield / Theoretical yield) × 10³ D) (Actual yield / Theoretical yield) × 100 % Yield = Actualyield/Theoreicayield×100 58 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 58) The number of moles of CO₂ which contains 16g of oxygen is KMU-CAT 2023 A) 0.25 B) 0.75 C) 0.5 D) 1 16g O = 1 mole O atoms. Since CO₂ has 2 O atoms/molecule, moles of CO₂ = 0.5. 59 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 59) Heating 24.8g of copper carbonate (CuCO₃) in a crucible produced only 13.9g of copper oxide (CuO). What is the percentage yield of copper oxide? KP-MDCAT 2024 A) 89.68% B) 83.98% C) 81.79% D) 86.87% Theoretical yield: CuCO₃ → CuO + CO₂. Molar masses: CuCO₃ = 123.5 g/mol, CuO = 79.5 g/mol. 24.8g CuCO₃ should yield (79.5/123.5)×24.8 = 15.96g CuO. % yield = (13.9/15.96)×100 = 87.09% (closest to option C) 60 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 60) How many moles of NaCl are produced from 16.5g of HCl, according to the neutralization reaction? HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) KP-MDCAT 2024 A) 0.252 B) 0.652 C) 0.852 D) 0.452 Moles HCl = 16.5g/36.5g/mol = 0.452 mol; 1:1 ratio gives 0.452 mol NaCl 61 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 61) The number of molecules in 8.96 L of a gas at 0°C and 1 atmosphere pressure is approximately A) 18.06×10²³ B) 24.08×10²² C) 12.04×10²³ D) 6.02×10²³ 22.4L of a gas at STP has no. of molecules 6 x1023, 8.96L of a gas at STP has no. of molecules 6 x1023 x 8.96 / 22.4 = 24.08 x1022 62 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 62) The number of water molecules in 1 litre of water is A) 18 B) 55.55 NA C) NA D) 18×1000 18g = NA = 18 ml, 1000 ml = NA x 1000/18 = 55.55 NA 63 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 63) How many atoms are contained in one mole of sucrose (C₁₂H₂₂O₁₁)? A) 5 × 6.023×10²³ atoms/mole B) None of these C) 5 × 6.023×10²³ atoms/mole D) 45 × 6.023×10²³ atoms/mole 1 mole of sucrose contains 6.023 x10 23 molecules, 1 molecule of sucrose has 45 atoms, 6.023 x10 23 molecule of sucrose has, 45 x 6.023 x10 23 atoms/mole 64 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 64) 2KClO₃ → 2KCl + 3O₂. Molecular mass of KClO₃ = 122.5 g/mol. For the production of 33.6 dm³ of O₂ at STP, the mass of KClO₃ to be decomposed is: ETEA 2018 A) 122.5g B) 245.0g C) 61.25g D) 367.5g 65 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 65) 2g of oxygen contains number of atoms equal to that in A) 2.3g of sodium B) 4g of sulphur C) 7g of nitrogen D) 0.5g of hydrogen 2gm of oxygen contains atom = 2/16 also 4g of sulphur = 4/32 = 2/16 66 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 66) The number of atoms in 4.25 g of NH₃ is approximately A) 2 × 10²³ B) 4 × 10²³ C) 1 × 10²³ D) 6 × 10²³ 1 mol = 17 g = NA Particles., 4 g = NA/4 particle NH3., Total atoms = NA/4 x 4= NA 67 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 67) Oxygen can be prepared by the decomposition of potassium chlorate (KClO₃). How many moles of oxygen O₂(g) can be formed by taking 12 moles of potassium chlorate (KClO₃) according to the following equation? 2KClO₃(s) + heat → 2KCl(s) + 3O₂(g) ETEA – 2023 A) 18 moles of oxygen B) 21 moles of oxygen C) 12 moles of oxygen D) 15 moles of oxygen 2KCIO3(1) + heat → 2KCl (s) + 302(g), 2 → 2;3, 2→3, 12→18 moles 68 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 68) How many molecules are present in 0.20 g of Hydrogen gas? ETEA 2013 A) 1.008/0.70 × 6.02×10²³ B) 0.20 × 2.016 C) 0.20/2.016 × 6.02×10²³ D) 0.20/1.008 × 6.02×10²³ For number of particles = N = n × NA, n = m/M=0.2/2.016 , N= 0.2/2.016×6.02 ×〖10〗^23 69 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 69) Which one of the following contains the greatest number of atoms: ETEA 2008 A) 71g of chlorine B) 4g of magnesium C) 127g of iodine D) 4g of Hydrogen For H2 = n = m/M=4/1.008=4 Mole , For = Mg = n = m/M=4/24=0.16 70 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 70) The number of molecules in 4.25 g of ammonia are A) 1.8×10²³ B) 3.5×10²³ C) 0.5×10²³ D) 1.5×10²³ Molecular weight of NH3 is 17 According to the mole concept, 17 gm NH3 has molecules 6 x1023, 1 gm NH3 has molecules = 6 x1023/17, 4.25 gm NH3 has molecules = 6 x1023x4.25/17 = 1.5 x1023 molecule 71 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 71) How many hydrogen atoms are present in one mole of water? ETEA 2012 A) 1.806 × 10²⁴ atoms B) 3.01 × 10²³ atoms C) 6.02 × 10²³ atoms D) 1.204 × 10²⁴ atoms To find # ofAtom= n × NA × # of Atom in formula = 1 × 6.02 × 1023 × 2 = 12.04× 1023 = 1.2 × 1024 atoms 72 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 72) Number of molecules in 100 ml of each of O₂, NH₃ and CO₂ at STP are A) NH₃ = CO₂ < O₂ B) CO₂ < O₂ < NH₃ C) NH₃ < O₂ < CO₂ D) NH₃ = CO₂ = O₂ According to avogadro’s hypothesis equal volumes of all gases under similar conditions of temperature and pressure contains equal no. of molecules. 73 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 73) The number of moles of sodium oxide in 620g of it is A) 10 moles B) 1 mol C) 100 moles D) 18 moles Sodium oxide → Na2 O, Molecular weight = 46 +16 = 62, 62gm of Na2 O = 1 mole, , 620gm of Na2 O = 10 mole. 74 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 74) Actual yield will reach the ideal (theoretical) value if the % yield of the reaction is, KP-MDCAT 2024 A) 100% B) 90% C) 10% D) 50% 100% yield means actual yield equals theoretical yield 75 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 75) 4.4 g of an unknown gas occupies 2.24L of volume at standard temperature and pressure. The gas may be A) Oxygen B) Carbon monoxide C) Sulphur dioxide D) Carbon dioxide 2.24L of gas has mass = 4.4gm, 22.4L of gas has mass 4.4 x 22.4 / 2.24 = 44, So given gas is CO2 because CO2 has molecular mass=44. 76 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 76) The efficiency of a chemical reaction can be expressed as: ETEA – 2020 A) Theoretical yield B) Maximum yield C) % yield D) Actual yield The efficiency of a chemical reaction is often expressed as the percentage yield, which is the ratio of the actual yield to the theoretical yield, multiplied by 100. 77 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 77) Four moles of electrons (4 × 6.02 × 10²³ electrons) would electroplate how many grams of silver from a silver nitrate solution? ETEA 2008 A) 216 B) 324 C) 540 D) 432 AgNO3 –––> Ag+3 + NO-3 (Ag = 108) , n = m/M,m=n×M, m = 4 × 108 = 432 78 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 78) The total number of gm-molecules of SO₂Cl₂ in 13.5g of sulphuryl chloride is A) 0.4 B) 0.3 C) 0.2 D) 0.1 Molecular weight of SO2Cl2 = 32 + 32 + 2 x 35.5 = 135gm, 135 gm of SO2Cl2= 1gm molecule, 13.5gm of SO2Cl2 = 13.5/135 = 0.1 79 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 79) One litre of a gas at STP weighs 1.16 g it can possibly be A) CO B) CH₄ C) C₂H₂ D) O₂ 1L of gas at S.T.P. weight 1.16g, 22.4 L of gas at S.T.P. weight =22.4x 1.16 = 25.984 = 26, This molecular weight indicates that given compound is C2H2 80 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 80) Natural chlorine occurs as a mixture of isotopes. If a mixture contains 75% Cl³⁵ and 25% Cl³⁷, what will be its correct atomic weight? ETEA 2010 A) 34.5 B) 35.5 C) 72 D) 70 Amount of Cl35 = 75/100=0.75,, Amount of Cl37 = 25/100=0.25 , Average atomic weight = (Amount) (At: Mass of 1st isotope) + (Amount) (At mass of 2nd isotope) = (0.75) (35) + (0.25)(37) = 26.25 + 9.25 = 35.5 Your score is 0% Restart quiz
