Stoichiometry and Chemical Calculations MCQs 2026

50 questions with detailed answers · 18 from past papers · 5 quiz batches available

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Page 1 of 1Questions 110 of 50
  1. Q1easy

    One mole of any substance contains ___ particles

    1. A6.022×10^22
    2. B3.011×10^23
    3. C1×10^23
    4. D6.022×10^23
    💡 Explanation:

    Avogadro's number, 6.022×10^23, defines the number of particles in one mole.

  2. Q2Past Paper · PPSC/FPSC/NTSeasy

    The molar mass of water (H2O) is

    1. A16 g/mol
    2. B20 g/mol
    3. C18 g/mol
    4. D17 g/mol
    💡 Explanation:

    Water's molar mass is (2×1) + 16 = 18 g/mol.

  3. Q3Past Paper · PPSC/FPSC/NTSeasy

    The number of moles in 44 g of CO2 (molar mass 44 g/mol) is

    1. A0.5 mol
    2. B2 mol
    3. C4 mol
    4. D1 mol
    💡 Explanation:

    Moles = mass ÷ molar mass = 44/44 = 1 mole.

  4. Q4easy

    The mass of 2 moles of NaOH (molar mass 40 g/mol) is

    1. A80 g
    2. B40 g
    3. C20 g
    4. D100 g
    💡 Explanation:

    Mass = moles × molar mass = 2 × 40 = 80 g.

  5. Q5Past Paper · PPSC/FPSC/NTSeasy

    The number of atoms in 1 mole of helium gas is

    1. A3.011×10^23
    2. B12.044×10^23
    3. C1×10^23
    4. D6.022×10^23
    💡 Explanation:

    One mole of any element contains Avogadro's number of atoms, 6.022×10^23.

  6. Q6easy

    Avogadro's number is approximately

    1. A3.14×10^23
    2. B6.022×10^23
    3. C9.8×10^23
    4. D1.6×10^-19
    💡 Explanation:

    Avogadro's number is defined as 6.022×10^23 particles per mole.

  7. Q7Past Paper · PPSC/FPSC/NTSeasy

    The volume occupied by 1 mole of an ideal gas at STP is

    1. A24 L
    2. B25 L
    3. C20 L
    4. D22.4 L
    💡 Explanation:

    At standard temperature and pressure, one mole of ideal gas occupies 22.4 liters.

  8. Q8easy

    The empirical formula represents

    1. AThe simplest whole-number ratio of atoms in a compound
    2. BThe exact number of atoms in one molecule
    3. CThe molar mass of the compound
    4. DThe number of moles present
    💡 Explanation:

    Empirical formulas give the simplest ratio of elements, not the actual molecular composition.

  9. Q9Past Paper · PPSC/FPSC/NTSeasy

    The molecular formula of a compound is always

    1. AA whole-number multiple of the empirical formula
    2. BSmaller than the empirical formula
    3. CUnrelated to the empirical formula
    4. DEqual to half the empirical formula
    💡 Explanation:

    Molecular formula equals the empirical formula multiplied by a whole number, including 1.

  10. Q10easy

    If the empirical formula of a compound is CH2O and its molar mass is 180 g/mol, its molecular formula is

    1. AC2H4O2
    2. BC3H6O3
    3. CC6H12O6
    4. DCH2O
    💡 Explanation:

    Empirical formula mass is 30; 180÷30=6, so the molecular formula is (CH2O)×6 = C6H12O6.

  11. Q11medium

    The percentage composition of an element in a compound is calculated as

    1. A(Mass of element in one mole ÷ Molar mass of compound) × 100
    2. B(Moles of compound ÷ Avogadro's number) × 100
    3. CAtomic mass ÷ 100
    4. DMolar mass ÷ mass of element
    💡 Explanation:

    Percent composition compares the mass contributed by an element to the total molar mass.

  12. Q12Past Paper · PPSC/FPSC/NTSmedium

    The percentage of oxygen by mass in water (H2O, molar mass 18) is approximately

    1. A88.9%
    2. B11.1%
    3. C50%
    4. D66.7%
    💡 Explanation:

    Oxygen contributes 16 out of 18 g/mol, giving 16/18 × 100 ≈ 88.9%.

  13. Q13medium

    The limiting reagent in a reaction is the reactant that

    1. AIs present in excess
    2. BDoes not react
    3. CIs completely consumed first, limiting the amount of product
    4. DHas the highest molar mass
    💡 Explanation:

    The reaction stops producing more product once the limiting reagent is used up.

  14. Q14Past Paper · PPSC/FPSC/NTSmedium

    The reactant that remains after a reaction is complete is called the

    1. AExcess reagent
    2. BLimiting reagent
    3. CCatalyst
    4. DProduct
    💡 Explanation:

    Any reactant left over after the limiting reagent is consumed is called the excess reagent.

  15. Q15medium

    Theoretical yield refers to

    1. AThe actual amount of product obtained in the lab
    2. BThe percentage of reactant used
    3. CThe maximum amount of product predicted by stoichiometric calculation
    4. DThe amount of limiting reagent
    💡 Explanation:

    Theoretical yield is the calculated maximum product possible assuming complete reaction.

  16. Q16medium

    Percentage yield is calculated as

    1. ATheoretical yield ÷ molar mass × 100
    2. B(Actual yield ÷ Theoretical yield) × 100
    3. C(Theoretical yield ÷ Actual yield) × 100
    4. DActual yield × molar mass
    💡 Explanation:

    Percentage yield compares the actual product obtained to the theoretically possible amount.

  17. Q17Past Paper · PPSC/FPSC/NTSmedium

    A balanced chemical equation obeys the law of

    1. ADefinite proportions only
    2. BMultiple proportions only
    3. CAvogadro's law
    4. DConservation of mass
    💡 Explanation:

    Balancing ensures equal numbers of atoms on both sides, honoring conservation of mass.

  18. Q18medium

    In the reaction N2 + 3H2 → 2NH3, the mole ratio of H2 to NH3 is

    1. A1:1
    2. B1:2
    3. C2:3
    4. D3:2
    💡 Explanation:

    The balanced equation shows 3 moles of H2 react to form 2 moles of NH3, a 3:2 ratio.

  19. Q19Past Paper · PPSC/FPSC/NTShard

    The formula weight of MgO (Mg=24, O=16) is

    1. A32
    2. B40
    3. C56
    4. D48
    💡 Explanation:

    Sum of atomic masses: 24+16 = 40.

  20. Q20medium

    The number of moles of oxygen atoms in 2 moles of CO2 is

    1. A4 mol
    2. B2 mol
    3. C1 mol
    4. D6 mol
    💡 Explanation:

    Each CO2 molecule has 2 oxygen atoms, so 2 moles of CO2 contain 4 moles of oxygen atoms.

  21. Q21Past Paper · PPSC/FPSC/NTSmedium

    Molarity of a solution is defined as

    1. AMoles of solute per kg of solvent
    2. BMoles of solute per liter of solution
    3. CGrams of solute per liter of solution
    4. DMoles of solvent per liter of solution
    💡 Explanation:

    Molarity (M) is defined as moles of solute divided by liters of total solution.

  22. Q22medium

    The molarity of a solution containing 2 moles of NaCl in 500 mL of solution is

    1. A1 M
    2. B0.5 M
    3. C2 M
    4. D4 M
    💡 Explanation:

    Molarity = 2 mol ÷ 0.5 L = 4 M.

  23. Q23medium

    Molality of a solution is defined as

    1. AMoles of solute per liter of solution
    2. BGrams of solute per liter of solvent
    3. CMoles of solvent per kg of solute
    4. DMoles of solute per kilogram of solvent
    💡 Explanation:

    Molality (m) is moles of solute per kilogram of solvent, independent of temperature-based volume changes.

  24. Q24Past Paper · PPSC/FPSC/NTSmedium

    The number of moles of solute in 250 mL of a 2 M solution is

    1. A1 mol
    2. B0.5 mol
    3. C2 mol
    4. D0.25 mol
    💡 Explanation:

    Moles = molarity × volume(L) = 2 × 0.25 = 0.5 mol.

  25. Q25medium

    The mass of 0.25 mol of CaCO3 (molar mass 100 g/mol) is

    1. A100 g
    2. B4 g
    3. C25 g
    4. D50 g
    💡 Explanation:

    Mass = moles × molar mass = 0.25 × 100 = 25 g.

  26. Q26medium

    The molar mass of glucose, C6H12O6, is

    1. A180 g/mol
    2. B162 g/mol
    3. C194 g/mol
    4. D176 g/mol
    💡 Explanation:

    Summing atomic masses: (6×12)+(12×1)+(6×16) = 72+12+96 = 180 g/mol.

  27. Q27Past Paper · PPSC/FPSC/NTSmedium

    The number of moles present in 11.2 L of a gas at STP is

    1. A1 mol
    2. B2 mol
    3. C0.25 mol
    4. D0.5 mol
    💡 Explanation:

    Moles = volume ÷ 22.4 L/mol = 11.2/22.4 = 0.5 mol.

  28. Q28medium

    According to the law of conservation of mass, in a chemical reaction the total mass of reactants

    1. AIncreases
    2. BEquals the total mass of products
    3. CDecreases
    4. DBecomes zero
    💡 Explanation:

    Mass is neither created nor destroyed, so total reactant mass equals total product mass.

  29. Q29medium

    The law of definite proportions states that a chemical compound always contains the same elements

    1. AIn variable proportions by mass
    2. BIn fixed proportions by mass
    3. CIn equal number of moles only
    4. DOnly in the gaseous state
    💡 Explanation:

    A pure compound always has the same elements combined in the same fixed mass ratio.

  30. Q30Past Paper · PPSC/FPSC/NTSmedium

    The law of multiple proportions applies when two elements form

    1. AOnly one compound
    2. BIonic compounds only
    3. CMore than one compound, with masses in small whole-number ratios
    4. DIsotopes
    💡 Explanation:

    When two elements form multiple compounds, the masses of one element combining with a fixed mass of the other form small whole-number ratios.

  31. Q31medium

    The number of moles of NaOH needed to completely neutralize 1 mole of H2SO4 is

    1. A1 mol
    2. B0.5 mol
    3. C2 mol
    4. D3 mol
    💡 Explanation:

    H2SO4 is diprotic, requiring 2 moles of NaOH for complete neutralization.

  32. Q32medium

    In the reaction 2H2 + O2 → 2H2O, the number of moles of water produced from 4 moles of H2 (excess O2) is

    1. A2 mol
    2. B1 mol
    3. C4 mol
    4. D8 mol
    💡 Explanation:

    The 2:2 mole ratio of H2 to H2O means 4 moles of H2 produce 4 moles of water.

  33. Q33Past Paper · PPSC/FPSC/NTSmedium

    A 1 molar solution of a substance contains how many moles of solute per liter

    1. A0.5 mol
    2. B1 mol
    3. C2 mol
    4. D10 mol
    💡 Explanation:

    By definition, a 1 M solution contains exactly 1 mole of solute per liter of solution.

  34. Q34medium

    Which quantity remains constant during a chemical reaction, per the law of conservation of mass

    1. ATotal mass
    2. BVolume of gas only
    3. CNumber of moles of gas
    4. DTemperature
    💡 Explanation:

    Total mass of the reacting system stays constant throughout a chemical reaction.

  35. Q35medium

    The formula mass of Na2CO3 (Na=23, C=12, O=16) is

    1. A84
    2. B106
    3. C90
    4. D62
    💡 Explanation:

    Sum: (2×23)+12+(3×16) = 46+12+48 = 106.

  36. Q36Past Paper · PPSC/FPSC/NTSmedium

    The percentage of carbon by mass in CO2 (molar mass 44) is approximately

    1. A50%
    2. B12%
    3. C27.3%
    4. D73%
    💡 Explanation:

    Carbon contributes 12 out of 44 g/mol, giving 12/44 × 100 ≈ 27.3%.

  37. Q37medium

    In stoichiometric calculations, mole ratios are obtained directly from the

    1. AAtomic masses
    2. BCoefficients of the balanced equation
    3. CMolar volumes
    4. DPercentage composition
    💡 Explanation:

    The coefficients in a balanced equation give the exact mole ratio between reactants and products.

  38. Q38medium

    If 5 moles of a limiting reagent produce 5 moles of product in a 1:1 reaction and only 4 moles of product were actually obtained, the percentage yield is

    1. A100%
    2. B125%
    3. C80%
    4. D50%
    💡 Explanation:

    Percentage yield = (4/5) × 100 = 80%.

  39. Q39Past Paper · PPSC/FPSC/NTSmedium

    A solution containing 1 mole of solute dissolved in exactly 1 kg of solvent has a molality of

    1. A0.5 m
    2. B1 m
    3. C2 m
    4. D10 m
    💡 Explanation:

    By definition, molality equals moles of solute per kilogram of solvent, giving 1 m here.

  40. Q40medium

    The relative formula mass of a compound is the sum of the

    1. AAtomic numbers of all atoms
    2. BNumber of moles of each element
    3. CCharges of all ions
    4. DRelative atomic masses of all atoms in the formula
    💡 Explanation:

    Formula mass is calculated by adding up the atomic masses of every atom shown in the formula.

  41. Q41medium

    Which of the following best describes an empirical formula determination experiment

    1. AMeasuring the boiling point of a compound
    2. BMeasuring the volume of gas released
    3. CTitrating an acid with a base
    4. DFinding the mass ratio of elements combined in a compound
    💡 Explanation:

    Empirical formula experiments typically measure the mass of each element present to find their simplest ratio.

  42. Q42Past Paper · PPSC/FPSC/NTShard

    The number of moles of ions produced when 1 mole of Na2SO4 fully dissociates in water is

    1. A1 mol
    2. B2 mol
    3. C4 mol
    4. D3 mol
    💡 Explanation:

    Na2SO4 dissociates into 2 Na+ ions and 1 SO4^2- ion, totaling 3 moles of ions.

  43. Q43hard

    The mass percent of hydrogen in methane, CH4 (molar mass 16 g/mol) is approximately

    1. A12.5%
    2. B20%
    3. C25%
    4. D75%
    💡 Explanation:

    Hydrogen contributes 4 out of 16 g/mol, giving 4/16 × 100 = 25%.

  44. Q44hard

    1 mole of a diatomic gas at STP occupies a volume of

    1. A22.4 L
    2. B11.2 L
    3. C44.8 L
    4. D6.022 L
    💡 Explanation:

    Molar volume at STP is 22.4 L for any ideal gas, regardless of whether it is diatomic.

  45. Q45hard

    How many moles of oxygen gas (O2) are required to completely combust 1 mole of methane (CH4 + 2O2 → CO2 + 2H2O)

    1. A1 mol
    2. B2 mol
    3. C3 mol
    4. D0.5 mol
    💡 Explanation:

    The balanced equation shows 2 moles of O2 are needed per mole of methane burned.

  46. Q46hard

    The number of grams in 0.5 mole of oxygen gas, O2 (molar mass 32 g/mol) is

    1. A32 g
    2. B16 g
    3. C8 g
    4. D64 g
    💡 Explanation:

    Mass = moles × molar mass = 0.5 × 32 = 16 g.

  47. Q47Past Paper · PPSC/FPSC/NTShard

    Gram atomic mass is the mass, in grams, of

    1. AOne mole of atoms of an element
    2. BOne atom of an element
    3. COne molecule of a compound
    4. DOne ion of an element
    💡 Explanation:

    Gram atomic mass equals the atomic mass expressed in grams for one mole of atoms.

  48. Q48hard

    The number of moles of electrons required to reduce 1 mole of Al3+ to Al metal is

    1. A3 mol
    2. B1 mol
    3. C2 mol
    4. D6 mol
    💡 Explanation:

    Al3+ requires gaining 3 electrons per ion to become neutral aluminum metal.

  49. Q49hard

    Stoichiometry is fundamentally based on the relationship between

    1. ATemperature and pressure of gases only
    2. BEnthalpy and entropy
    3. CAtomic radius and ionization energy
    4. DMoles of reactants and products in a balanced equation
    💡 Explanation:

    Stoichiometric calculations use the mole ratios from a balanced chemical equation.

  50. Q50hard

    The number of moles in 9 g of water (molar mass 18 g/mol) is

    1. A1 mol
    2. B0.25 mol
    3. C0.5 mol
    4. D2 mol
    💡 Explanation:

    Moles = mass ÷ molar mass = 9/18 = 0.5 mol.