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) 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) 367.5g B) 122.5g C) 245.0g D) 61.25g 2 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 2) How many mole of helium gas occupy 22.4 L at 0°C at 1 atm. pressure? A) 0.9 B) 1 C) 0.11 D) 1.11 1mole of any gas at STP occupies 22.4L. 3 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 3) 60 a.m.u of C-12 contain carbon: atoms ETEA 2016 A) 5 B) 60 × 6.02 × 10²³ C) 60 D) 5 × 6.02 × 10²³ number of atoms = 60/12 = 5 atom s 4 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 4) How many hydrogen atoms are present in one mole of water? ETEA 2012 A) 6.02 × 10²³ atoms B) 3.01 × 10²³ atoms C) 1.204 × 10²⁴ atoms D) 1.806 × 10²⁴ atoms To find # ofAtom= n × NA × # of Atom in formula = 1 × 6.02 × 1023 × 2 = 12.04× 1023 = 1.2 × 1024 atoms 5 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 5) 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) NH₃ B) HN₂ C) NH₂ 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. 6 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 6) 2g of oxygen contains number of atoms equal to that in A) 4g of sulphur B) 2.3g of sodium C) 7g of nitrogen D) 0.5g of hydrogen 2gm of oxygen contains atom = 2/16 also 4g of sulphur = 4/32 = 2/16 7 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 7) The number of molecules of CO₂ present in 44g of CO₂ is A) 3.023×10¹⁰ B) 12.023×10²³ C) 6.023×10²³ D) 3.023×10²³ 1 mol CO2 = 44g = 6.023 x10 23 molecules 8 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 8) 19.7 kg of gold was recovered from a smuggler. How many atoms of gold were recovered (Au=197)? A) 6.023×10²⁵ B) 6.023×10²⁴ C) 100 D) 6.023×10²³ 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. 9 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 9) Efficiency of chemical reaction can be checked by calculating KP-MDCAT 2024 A) Theoretical yield B) Percentage yield C) Actual yield D) Amount of the reactant unused Percentage yield compares actual yield to theoretical yield, showing reaction efficiency 10 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 10) The efficiency of a chemical reaction can be expressed as: MDCAT 2020 A) % yield B) Maximum yield C) Actual yield D) Theoretical 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 11 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 11) One mole is the amount of substance which contains as many elementary entities as contained in: ETEA 2012 A) 0.12 kg of ⁶C₁₂ B) 1.2 kg of ⁶C₁₂ atom C) 0.012 kg of ⁶C₁₂ atom D) 0.12 kg of ⁸O₁₆ One mole of C-12 = 12 g = (0.012 Kg) 12 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 12) The numbers of moles of BaCO₃ which contain 1.5 moles of oxygen atoms is A) 1 B) 6.023×10²³ C) 0.5 D) 3 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 13 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 13) Theoretical yield is always: ETEA 2015 A) None of the above B) Greater than actual yield C) Both are equal D) Less than practical 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. 14 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 14) Which of the following pairs have the same electronic structure? ETEA 2006 A) Ag & Sn B) Ca & Ar C) Ar & Cl D) Mg & Na Ar &Cl are Iso-electronic between both have 18 e. 15 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 15) Which one of the following contains the greatest number of atoms: ETEA 2008 A) 4g of magnesium B) 127g of iodine C) 71g of chlorine D) 4g of Hydrogen For H2 = n = m/M=4/1.008=4 Mole , For = Mg = n = m/M=4/24=0.16 16 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Limiting and excess reagent 16) 2XeF₆ + SiO₂ → 2XeOF₄ + SiF₄ Consider the above chemical reaction. If 122.6 g of XeF₆ reacts with 60 g of SiO₂ to form the products. Select the limiting reagent and amount of SiF₄ formed: (XeF₆ = 245.3 amu, SiO₂ = 60 amu, SiF₄ = 104 amu) ETEA 2016 A) SiO₂, 52 g B) XeF₆, 26 g C) SiO₂, 26 g D) XeF₆, 52 g Moles of XeF6=122.6/245.3=0.5 , Moles of SiO2=60/60=1.0 mol, Step 2: Determine the limiting reactant. The reaction requires 2mol of XeF6 to react with 1mol of SiO2., For 0.5mol of XeF6, we need 0.25mol of SiO2., We have 1.0mol ofSiO2, which is more than enough., Thus, XeF6 is the limiting reactant., 2mol of XeF6 produces 1mol of SiF4., 0.5mol of XeF6 produces0.25mol of SiF4., Molar mass of SiF4=104g/mol., Mass of SiF4=0.25×104=26 g 17 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 17) A gas at STP contains only 6.023×10²³ atoms and is monatomic it will occupy. ETEA 2010 A) 30.5L B) 44.8L C) 1.2L D) 22.4L 22.4dm3 = 22.4L = 6.023×1023 atoms 18 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 18) 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) 2 × 6.023 × 10²³ C) 6.023 × 10²³ D) None of the above 4 x NA = 4 x 6.023×1023 19 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 19) What are the Avogadro’s number of particles in 0.25 moles of CO₂? NUMS 2020 A) 6.022 × 10⁷ B) 2.00 × 10²³ C) 1.505 × 10¹⁵ D) 1.505 × 10²³ 1 mol = NA → 0.25 mol = NA /0.25 = 1.505 × 1023 20 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 20) One mole of which of the following bucky ball will have more molecules? ETEA – 2019 A) all same B) C₅₀ C) C₆₀ D) C₂₀ One mole of any substance, including buckyballs, contains Avogadro’s number of molecules, so they all have the same number of molecules 21 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 21) What is the mass of 1 mole of calcium carbonate (CaCO₃)? KP-MDCAT 2024 A) 50g B) 75g C) 100g D) 125g Molar mass of CaCO₃ = 40 + 12 + (3×16) = 100 g/mol 22 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 22) A sample of carbon-12 has a mass of 3.0 g. Which expression gives the number of atoms in the sample? (NA is the symbol for the Avogadro constant). ETEA 2013 A) 0.25NA B) 3.0NA C) 4.0NA D) 0.0030NA N = n × NA = n = m/M=3/12=0.25 , N = 0.25 × NA 23 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 23) 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) 432 D) 540 AgNO3 –––> Ag+3 + NO-3 (Ag = 108) , n = m/M,m=n×M, m = 4 × 108 = 432 24 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Limiting and excess reagent 24) 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) H₂ B) N₂ C) Both a & b 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. 25 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 25) 1 amu is equal to A) 1/12 of C-12 B) 1.66 × 10⁻¹³ kg C) 1g of H D) 1/14 of O-16 1 amu is equal to = 1/12 of C- 12 26 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 26) Identify the correct option with same empirical formula for both compounds NUMS 2020 A) H₂O & H₂O₂ B) C₆H₁₂ & C₆H₆ C) H₂S₂O₃ & H₂SO₄ D) C₆H₁₂O₆ & CH₃COOH (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 27 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 27) How many grams of CO₂ can be produced by thermally decomposing 10 moles of ZnCO₃(S)? KP-MDCAT 2024 A) 360 B) 440 C) 320 D) 400 ZnCO₃ → ZnO + CO₂. 10 moles ZnCO₃ produces 10 moles CO₂ = 10 × 44g = 440g 28 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 28) What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene? A) 2.8 kg B) 6.4 kg C) 96 kg D) 9.6 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. 29 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 29) 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) 72 B) 70 C) 34.5 D) 35.5 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 30 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 30) The molecular weight of a gas is 45. Its density at STP is A) 22.4 B) 2 C) 11.2 D) 5.7 The density of gas = molecular wt. of metal / volume = 45/22.4 = 2gm/L 31 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 31) Many elements have fractional atomic masses. This is because NUMS 2020 A) Atomic masses are average masses of isotopes proportional to their abundance B) Atomic masses are average masses of isotopes C) Mass of atom is itself fractional D) Atomic masses are average masses of isobars 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. 32 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 32) Which contains more atoms? ETEA – 2019 A) 8 gram Na B) 9 gram Ag C) all same D) 7 gram Mg Greater the number of moles, greater the number of atoms. , Moles of Mg = 7g / 24g/mol = 0.2917, Moles of Na = 8g / 23g/mol = 0.3478, Moles of Ag = 9g / 108g/mol = 0.0833 33 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 33) 117.0g of NaCl have A) 1.204 × 10²⁴ formula units of NaCl B) 1.204 × 10²³ molecules of NaCl C) 12.04 × 10²² formula units 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 34 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 34) How many moles are there in 60g of sodium hydroxide (NaOH)? KP-MDCAT 2024 A) 8 B) 2 C) 6 D) 4 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) 35 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Limiting and excess reagent 35) The reactant which consumes first in a chemical reaction is called? ETEA – 2023 A) Excess reactant B) Consumed reactant C) Non-reacted reactant D) Limiting reactant The reactant which consumes first in a chemical reaction is called Limiting reactant 36 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 36) Cylinder “A” contains 4.6 grams of C₂H₅OH and cylinder “B” has 3 grams C₂H₆: ETEA 2016 A) Both cylinder A and B have equal number of molecules B) none of these C) Both cylinders have the equal number of hydrogen atoms D) Cylinder A has greater number of molecules than cylinder B 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. 37 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 37) A mixture of 10cm³ of oxygen and 50cm³ of hydrogen is sparked continuously. What is the maximum theoretical decrease in volume? ETEA – 2019 A) 30cm³ B) 15cm³ C) 20cm³ D) 10cm³ When hydrogen and oxygen react (2H2 + O2 → 2H2O), two volumes of water vapor are formed from two volumes of hydrogen and one volume of oxygen. Starting with 10 cm^3 of O2 and 50 cm^3 of H2, only 20 cm^3 of H2 can react with all 10 cm^3 of O2 based on stoichiometry, decreasing the volume by 30 cm^3. 38 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 38) One litre of a gas at STP weighs 1.16 g it can possibly be A) O₂ B) C₂H₂ C) CH₄ D) CO 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 39 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 39) The number of oxygen atoms in 4.4 g of CO₂ is approx. A) 12×10²³ B) 6×10²³ C) 6×10²³ D) 1.2×10²³ 44g of CO2 has 2 x 6 x 1023 atoms of oxygen., 4.4g of CO2 has 12 x 1023 x 4.4/44 = 1.2 x1023 40 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 40) The number of molecules in 4.25 g of ammonia are A) 1.5×10²³ B) 1.8×10²³ C) 0.5×10²³ D) 3.5×10²³ 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 41 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 41) How many oxygen atoms are present in 278g of Hydrated Ferrous Sulphate? (FeSO₄.7H₂O = 278 amu) ETEA 2016 A) 6.525 × 10²⁴ B) 2.408 × 10²³ C) 6.023 × 10²² D) 6.023 × 10²³ 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 42 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 42) Avogadro’s constant is the number of ETEA – 2019 A) Atoms in 1g of He B) atoms in 24g of Mg C) electrons present in 2g H 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. 43 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 43) The volume occupied by 3.2g of oxygen gas at S.T.P is: ETEA – 2023 A) 24 dm³ B) 22.4 dm³ C) 1.12 dm³ D) 2.24 dm³ 32 g = 22.8 dm3, .2 g = 2.28dm3 44 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 44) Number of gm of oxygen in 32.2 g Na₂SO₄.10H₂O is A) 2.08 B) 22.4 C) 2.24 D) 20.8 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. 45 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 45) Which of the following is a compound? ETEA 2007 A) O₂ B) Brass C) NH₃ D) Air Brass ––> Alloy (Made of Cu + Zn), O2 –––> Molecule 46 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 46) The number of molecules in 16 g of methane is A) 16/6.023×10²³ B) 6.023×10²³ C) 3×10²³ D) 16/3×10²³ 16 g = 1 mol = 6.023 x10 23 47 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 47) Which compound with the given information has greater mass in Kg? ETEA 2018 A) 6.02 × 10²³ molecules of C₂H₄ B) All have equal mass C) 2 mole of CO D) 22.4 km³ N₂ at STP 22.4 km3 N2 at STP = 28 g = 2 mol CO 48 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 48) What mass of aluminium oxide (Al₂O₃) is produced from 18.5g of Al metal, when it reacts completely with oxygen gas according to the following equation? 4Al(s) + 3O₂(g) → 2Al₂O₃(s) KP-MDCAT 2024 A) 36.5g B) 34.9g C) 32.6g D) 30.8g 4Al → 2Al₂O₃; 18.5g Al = 0.685 mol → 0.3425 mol Al₂O₃ = 34.9g 49 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 49) The amount of products that is actually produced during a chemical reaction by performing an experiment is called _____ ETEA – 2022 A) Mole B) Percent yield C) Theoretical yield D) Actual 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. 50 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 50) Na is Iso-electronic with: ETEA 2016 A) Mg B) He C) Ne D) Fe Na+ = 10 ȇ, Ne = 10e –––>Both are Iso-electronic 51 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 51) The number of molecules in 8.96 L of a gas at 0°C and 1 atmosphere pressure is approximately A) 12.04×10²³ B) 24.08×10²² C) 6.02×10²³ D) 18.06×10²³ 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 52 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 52) 1 mol of CH₄ contains A) 3.0 g of carbon B) 1.81×10²³ molecules of CH₄ C) 6.02 × 10²³ atoms of H D) 4 g atom of Hydrogen 1 mol of CH4 contains 4 g atom of Hydrogen 53 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 53) How many atoms are contained in one mole of sucrose (C₁₂H₂₂O₁₁)? A) 45 × 6.023×10²³ atoms/mole B) 5 × 6.023×10²³ atoms/mole C) 5 × 6.023×10²³ atoms/mole D) None of these 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 54 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 54) 1 amu is equal to 1.661 × 10⁻²⁴g, then 1.0 g will be equal to: ETEA 2012 A) 6.022 × 10²⁴amu B) 6.022 × 10⁻²³amu C) 6.022 × 10⁻²⁴amu D) 6.022 × 10²³amu number of amu = 1/1.61 x 10-24 = 2.023×1023 55 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 55) The mass of a molecule of water is A) 2.5 × 10⁻²⁶ kg B) 3 × 10⁻²⁶ kg C) 1.5 × 10⁻²⁶ kg D) 3 × 10⁻²⁵ kg 6 x1023 molecules has mass = 18gm, 1 molecules has mass 18/6 x1023 = 3 x 10-23 gm 56 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 56) The largest number of molecules is in A) 46g of CH₃OH B) 28g of CO₂ C) 54g of N₂O₅ D) 34g of water 57 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 57) Avogadro’s number may represent KMU-CAT 2023 A) Number of particles B) Mass of particles C) All of the above D) Volume of particles 6.022×10²³ particles/mol. It does not directly represent mass (c) or volume (a). 58 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 58) Which one of the following pairs of gases contains the same number of molecules? A) 32 g of O₂ and 32 g of N₂ B) 16 g of O₂ and 14 g of N₂ C) 8 g of O₂ and 22 g of CO₂ D) 28 g of N₂ and 22 g of CO₂ 16 g of O2 = 16/32 = 0.5 mol, 14 g of N2 = 14/28 = 0.5 mol 59 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 59) 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 60 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 60) The number of gram atoms in 3g Hydrogen atoms is the same as the number of gram atoms in 48g of ETEA 2018 A) O₂ B) N C) O D) C 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 61 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 61) A ring contains 1.2 gram of diamond, the number of carbon atoms in the ring are: ETEA 2016 A) NA/2 B) 1.2NA C) NA/10 D) NA n = 1.2/12 = 0.1 mol →no of atoms = 0.1 x NA = NA/10 62 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 62) 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. 63 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 63) What is the number of hydrogen atoms in 5 moles of water? ETEA 2015 A) 3.0115 × 10²⁴ B) 6.023 × 10²³ C) 6.023 × 10²⁴ D) 5.0 × 10²³ # of Atoms = n × NA × # of Atoms in formula (Atomicity) = 5× 6.022 × 1023 × 2=6.023 × 1024 64 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 64) Which of the following contain the same number of molecules as 22 gram of carbon dioxide? ETEA – 2022 A) 9g of water B) 71g of chlorine gas C) 2g of hydrogen gas D) 32g of oxygen gas 22/44 = 0.5 moles CO2., 9/18 = 0.5 g water 65 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 65) Which of the following has least mass? A) 7.0 g of Ag B) 2 g atom of nitrogen C) 1 mole of S D) 3×10²³ atoms of C 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. 66 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 66) The number of electrons in a mole of hydrogen molecule is A) 6.023×10²³ B) 12.023×10²³ C) Indefinite D) 3.023×10²³ One hydrogen has 1 electron. So one mole will have 1 mol electrons. . 67 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 67) The volume occupied by 4.4 g of CO₂ at STP is A) 0.1 L B) 0.224 L C) 22.4 L D) 2.24 L 44g CO2 occupies 22.4L at STP, 4.4g CO2 occupies 22.4×4.4/44 = 2.24 L 68 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Actual and Theoretical yield 68) The efficiency of a chemical reaction can be expressed as: ETEA – 2020 A) % yield B) Theoretical yield C) Maximum 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. 69 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 69) The number of moles of sodium oxide in 620g of it is A) 1 mol B) 18 moles C) 100 moles D) 10 moles Sodium oxide → Na2 O, Molecular weight = 46 +16 = 62, 62gm of Na2 O = 1 mole, , 620gm of Na2 O = 10 mole. 70 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 70) One molecule of oxygen has the same mass as ETEA – 2023 A) One gram atom of Sulphur B) 2 molecules of CH₄ C) 16 moles of H₂ D) 8 moles of helium CH4 = 16 amu x 2 = 32 amu 71 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 71) Number of molecules in 100 ml of each of O₂, NH₃ and CO₂ at STP are A) CO₂ < O₂ < NH₃ B) NH₃ = CO₂ < O₂ C) NH₃ = CO₂ = O₂ D) NH₃ < O₂ < CO₂ According to avogadro’s hypothesis equal volumes of all gases under similar conditions of temperature and pressure contains equal no. of molecules. 72 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 72) 1 mole of any substance contains ____ substances NUMS 2020 A) 6.02 × 10²⁴ B) 6.02 × 10²² C) 6.02 × 10²³ D) 3.01 × 10²³ 1 mole of any substance contains 6.02 × 1023 73 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 73) The weight of 1×10²² molecules of CuSO₄.5H₂O is A) None of these B) 41.59 g C) 415.9 g D) 4.159 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 74 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 74) The number of moles of CO₂ which contain 8 gm of oxygen are NUMS 2020 A) 1 B) 4.5 C) 0.5 D) 0.25 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 75 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 75) Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85g of BaCO₃ (Atomic mass of Ba=137) will be A) 2.24 L B) 0.84 L C) 1.12 L D) 4.06 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. 76 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 76) The number of gram atoms of oxygen present in 0.3 gram mole of (COOH)₂.2H₂O A) 1.8 B) 1.2 C) 3.6 D) 0.6 1 mol (COOH)2.2H2O = 96 g, 0.3mol = 28.8 g, No. of gram atoms of oxygen = 28.8/16 = 1.8 77 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 77) 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) 86.87% B) 81.79% C) 89.68% D) 83.98% 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) 78 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 78) 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) 21 moles of oxygen B) 18 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 79 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> STOICHIOMETRY CALCULATION 79) 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) 0.452 B) 0.252 C) 0.652 D) 0.852 Moles HCl = 16.5g/36.5g/mol = 0.452 mol; 1:1 ratio gives 0.452 mol NaCl 80 / 80 Category: BOMVirtual.com >> Chemistry >> 01 Stoichiometry >> Mole 80) Total number of atoms represented by the compound CuSO₄.5H₂O is A) 27 B) 21 C) 5 D) 8 CuSO4.5H2O = 1+1+9+10=21 Your score is 0% Restart quiz
