Class 10 Chemistry Chapter 15 Stoichiometry Complete Unit Test 7
Grade 10 Chemistry Chapter 15 Board standard Model Paper
Prepare for your annual examination with the Class 10 Chemistry Chapter 15 Stoichiometry Complete Chapter Test 7, included in the Punjab Board 2027 Model Papers. This board-standard Chemistry paper covers the complete chapter and helps students practice important concepts, chemical equations, stoichiometric calculations, numerical problems, and exam-style questions. Attempt this Class 10 Chemistry Complete Chapter Test to assess your preparation, improve problem-solving skills, and prepare confidently for the Punjab Board Class 10 Chemistry examination 2027.
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Note: Four possible answers A, B, C and D to each question are given. The choice which you think is correct, fill that circle in front of that question with Marker or Pen ink. Cutting or filling two or more circles will result in zero mark in that question.
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What does the concentration of a solution measure?(a) The temperature of the solution(b) The amount of solvent only(c) The amount of solute dissolved in a solution(d) The pressure of the solution
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What is the unit for mass/volume concentration?(a) mol/cm3(b) g/dm3(c) kg/m3(d) mol/g
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Percentage composition of a compound refers to:(a) The mass of one mole of the compound(b) The percentage mass of each element in the compound(c) The ratio of atoms in a compound(d) The total mass of all elements
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The empirical formula of hydrogen peroxide (H2O2) is(a) H2O(b) HO(c) H2O2(d) H3O
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Why do ionic compounds not have molecular formulae?(a) They exist as molecules(b) They have variable composition(c) They exist as ions in a fixed ratio(d) They contain covalent bonds
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If the mole ratio of elements in a compound is 1 : 1.5, what should be done to obtain a whole number ratio?(a) Divide by 1.5(b) Multiply by 2(c) Multiply by 3(d) Divide by 2
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Why is one reactant often taken in excess during a chemical reaction?(a) To speed up the reaction(b) To increase temperature(c) To ensure complete use of the expensive reactant(d) To reduce side reactions
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Theoretical yield is:(a) Always greater than actual yield(b) Always equal to actual yield(c) Always less than actual yield(d) Unrelated to balanced equation
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Volume occupied by 15 moles of ammonia gas at RTP:(a) 224 dm3(b) 160 dm3(c) 360 dm3(d) 265 dm3
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What will the molar concentration of potassium hydroxide solution if 2.0 g of it are dissolved in 100 cm3 of solution?(a) 0.36 mol/dm3(b) 0.53 mol/dm3(c) 0.70 mol/dm3(d) 0.45 mol/dm3
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If the molar mass of a compound is 178 g/mol and its empirical formula is C3H5O3. What is its molecular formula?(a) C3H5O3(b) C6H10O6(c) C9H15O9(d) C8H10O8
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If the actual yield of a compound is 0.198 g while the theoretical field is 0.217 g. What will be the percentage yield?(a) 50%(b) 91.2%(c) 80%(d) 90%
Here is Subjective Part of Model paper.
By Avogadro’s law, equal volumes of gases contain equal numbers of molecules (moles). At RTP one mole of any gas occupies 24 dm3, but the molar masses of gases differ, so the masses of the same volume differ.
Volume = 500 cm3 = 0.5 dm3
Concentration = 1.8 g ÷ 0.5 dm3 = 3.6 g/dm3
It is the percentage by mass of each element present in a compound.
They are used to find the amounts of reactants needed and products obtained in industry, to work out medicine dosages, to prepare fertilizers and food products, to estimate cost and purity, and to avoid wastage.
Ionic compounds do not exist as discrete molecules. They form a crystal lattice of ions in a fixed ratio, so only the simplest ratio of ions (formula unit) can be written.
1) Find the mass (or percentage) of each element. 2) Divide each by its atomic mass to get moles. 3) Divide all by the smallest number of moles. 4) Convert to the simplest whole-number ratio (multiply if needed). 5) Write the formula with these numbers as subscripts.
A molecular formula shows the actual number of atoms of each element in one molecule. Examples: glucose C6H12O6, hydrogen peroxide H2O2, benzene C6H6.
The coefficients give the mole ratio. Convert the given mass to moles, use the mole ratio to get the moles of the required substance, then multiply by its molar mass to get the mass.
It lets the exact amount of the active ingredient be calculated, so that the correct dose is given without overdose or under-dose.
Moles of CaCO3 = 25 ÷ 100 = 0.25 mol
1 mol CaCO3 gives 1 mol CaCl2, so 0.25 mol CaCl2.
Mass = 0.25 × 111 = 27.75 g
The reactant that is completely used up first and so limits the amount of product is the limiting reactant. Importance: it decides the maximum product formed, and helps to save the costly reactant and avoid waste.
A balanced equation obeys the law of conservation of mass, and its coefficients give the correct mole ratios needed for the calculation.
The maximum amount of product that can be obtained from the given reactants, calculated from the balanced equation.
Percentage yield = (actual yield ÷ theoretical yield) × 100
Moles of Zn = 50 ÷ 65 = 0.77 mol
Molar mass of Zn(NO3)2 = 189 g/mol
Theoretical yield = 0.77 × 189 ≈ 145.4 g
Percentage yield = (80.2 ÷ 145.4) × 100 ≈ 55.2 %
It lets us convert between the volume and the moles (or mass) of a gas: one mole of any gas occupies 22.4 dm3 at STP (24 dm3 at RTP), so gas amounts can be found from volume without weighing.
Avogadro’s law says equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. So one mole (6.02 × 1023 molecules) of every gas occupies the same volume, the molar volume.
Because empirical formula shows only the simplest ratio of atoms. Different compounds can have the same ratio but different molecular formulae, e.g. CH2O (formaldehyde), C2H4O2 (acetic acid) and C6H12O6 (glucose).
It shows the efficiency of a reaction and helps industries judge the cost, the losses and the success of a process.
Molar mass of H2SO4 = 98 g/mol
Moles = 49 ÷ 98 = 0.5 mol
Concentration = 0.5 mol ÷ 1 dm3 = 0.5 mol/dm3
Convert the masses of the reactants to moles and compare them with the mole ratio of the balanced equation (or calculate the product from each reactant). The reactant giving the smaller amount of product is the limiting reactant.
Impurities reduce the amount of the active compound and can be harmful. Knowing the purity gives the correct dose and ensures safety and effectiveness.
No. A reversible reaction does not go to completion, so no reactant is completely used up; the reaction reaches equilibrium.
If the subscripts are in the simplest whole-number ratio with no common factor, the formula is empirical (e.g. CH2O). If they can be divided by a common factor (e.g. C2H4), it is not.
Molar volume is the volume occupied by one mole of a gas at a given temperature and pressure. It is 22.4 dm3 at STP and 24 dm3 at RTP for every gas.
Example: 1 mole of O2 (32 g) and 1 mole of H2 (2 g) each occupy 22.4 dm3 at STP; 2 moles of CO2 occupy 44.8 dm3.
Volume = 150 cm3 = 0.15 dm3
Moles = 0.4 × 0.15 = 0.06 mol
Mass = 0.06 × 58.5 = 3.51 g
1) Find the mass or percentage of each element. 2) Divide by atomic mass to get moles. 3) Divide all by the smallest number of moles. 4) Convert to a whole-number ratio. 5) Write the formula using these numbers as subscripts.
C = 40.0 ÷ 12 = 3.33; H = 6.71 ÷ 1 = 6.71; O = 53.3 ÷ 16 = 3.33
Dividing by 3.33 gives C : H : O = 1 : 2 : 1, so the empirical formula is CH2O (mass 30).
Molar mass = 28.8 ÷ 0.320 = 90 g/mol; n = 90 ÷ 30 = 3
Molecular formula = C3H6O3
Convert the masses of the reactants to moles, compare them with the mole ratio of the balanced equation, and find which reactant would be finished first (or gives less product). That reactant is the limiting reactant.
Moles of H2 = 12 ÷ 2 = 6 mol
Moles of N2 = 64 ÷ 28 = 2.29 mol
6 mol H2 needs 6 ÷ 3 = 2 mol N2 (available 2.29 mol), so N2 is in excess.
Limiting reactant = H2
