We are providing you with the best chemistry notes written by Sir Umair Khan. This is the 2nd chapter of Class 10 Chemistry, Chapter 15 Stoichiometry. All notes for class 10 chemistry are here, and All Video lectures are here. Watch and Subscribe, please.
A. Multiple Choice Questions (Chapter 15 Stoichiometry)
1. Volume occupied by 15 moles of ammonia gas at RTP:
(a) 224 dm3
(b) 160 dm3
(c) 360 dm3
(d) 265 dm3
Calculation: Volume = moles × 24 dm3/mol = 15 × 24 = 360 dm3.
2. How many moles of natural gas CH4, will be present in a cylinder if its volume is 18 dm3 at RTP?
(a) 0.75 mol
(b) 0.67 mol
(c) 0.85 mol
(d) 0.56 mol
Calculation: Moles = Volume / 24 = 18 / 24 = 0.75 mol.
3. What will be the molar concentration of the potassium hydroxide solution if 2.0 g of it is 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
Calculation: Molar mass of KOH = 39 + 16 + 1 = 56 g/mol. Moles = 2.0 / 56 = 0.0357 mol. Volume = 0.1 dm3. Concentration = 0.0357 / 0.1 = 0.357 ≈ 0.36 mol/dm3.
4. What will be the mass of NaCl needed to make up 15 dm3 of its solution with a concentration of 0.7 g/dm3?
(a) 21 g
(b) 10.5 g
(c) 31.5 g
(d) 5.4 g
Calculation: Mass = Concentration (g/dm3) × Volume (dm3) = 0.7 × 15 = 10.5 g.
5. Find out the concentration of H2SO4 in mol/dm3 if its 10 cm3 reacts with 20 cm3 of NaOH solution having concentration 0.1 mol/dm3.
(a) 0.01 mol/dm3
(b) 0.001 mol/dm3
(c) 0.1 mol/dm3
(d) 0.15 mol/dm3
Calculation: Equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O. Moles of NaOH = 0.020 × 0.1 = 0.002 mol. Moles of H2SO4 required = 0.002 / 2 = 0.001 mol. Concentration of H2SO4 = 0.001 mol / 0.010 dm3 = 0.1 mol/dm3.
6. 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) D) C8H10O8
Calculation: Empirical formula mass of C3H5O3 = (3×12) + (5×1) + (3×16) = 36 + 5 + 48 = 89 g/mol. n = Molar mass / Empirical mass = 178 / 89 = 2. Molecular formula = 2 × (C3H5O3) = C6H10O6.
7. When natural gas burns in air, which component is the limiting reactant?
(a) Both components
(b) Neither component is a limiting reactant
(c) Air
(d) Natural gas
Explanation: Air (containing oxygen) is present in an open abundance environment, making the fuel (natural gas) the limiting reactant that determines when combustion stops.
8. If the actual yield of a compound is 0.198 g while the theoretical yield is 0.217 g, what will be the percentage yield?
(a) 50%
(b) 91.2%
(c) 80%
(d) 90%
Calculation: Percentage Yield = (Actual Yield / Theoretical Yield) × 100 = (0.198 / 0.217) × 100 = 91.24% ≈ 91.2%.
B. Short Answer Questions
15.1. What is the molar volume of a gas at RTP?
Answer: Molar volume is the volume that one mole of any gas takes up. At Room Temperature and Pressure (RTP), this volume is always 24 dm3 (or 24 liters).
15.2. How is the concept of molar volume useful for us?
Answer: It helps us easily find out how many moles or grams of a gas we have just by measuring its volume. This means we do not need to weigh a gas on a scale, which is very hard to do.
15.3. How does the molar volume relate to Avogadro’s law?
Answer: Avogadro’s law says that equal volumes of all gases have the same number of molecules at the same temperature and pressure. Molar volume matches this because exactly 1 mole of any gas will always take up 24 dm3 at RTP.
15.4. Why two different compounds may show the same empirical formula?
Answer: An empirical formula only shows the simplest ratio of atoms, not the real total number of atoms. Different chemicals can have the exact same ratio of atoms but have different total amounts. For example, both acetylene (C2H2) and benzene (C6H6) simplify to the ratio CH.
15.5. Why is percentage yield important?
Answer: Percentage yield tells us how successful an experiment was. It shows how much product we actually made compared to the maximum amount we expected to make. A high percentage yield means very little material was wasted.
15.6. What is the concentration of a solution in mol/dm3 if it contains 49 g of H2SO4 in one dm3 of solution?
Answer:
1. Find the mass of 1 mole of H2SO4: (2 × 1) + 32 + (4 × 16) = 98 g/mol.
2. Find the number of moles: 49 g ÷ 98 g/mol = 0.5 moles.
3. Find concentration: 0.5 moles ÷ 1 dm3 = 0.5 mol/dm3.
C. Constructed Response Questions (Chapter 15 Stoichiometry)
15.1. How would you identify the limiting reactant in a chemical reaction?
Answer: To find the limiting reactant:
1. Change the starting amounts of all reactants into moles.
2. Look at the balanced chemical equation to see how much of each reactant is needed.
3. The reactant that finishes first and stops the reaction is the limiting reactant.
15.2. Why is it important to find the purity of a compound which is used as a medicine?
Answer: Medicines must be pure because harmful impurities can make a patient sicker or cause unexpected side effects. Knowing the purity also ensures that the patient gets the exact, correct dose of the drug to cure them safely.
15.3. Will there be a limiting reactant in a reversible reaction?
Answer: No, there is no limiting reactant in a reversible reaction. This is because the reaction never completely finishes. It moves forward and backward at the same time, so none of the reactants are completely used up.
15.4. How do you know if a formula is empirical or not?
Answer: Look at the numbers (subscripts) in the chemical formula. If you cannot divide them any further by the same number, it is an empirical formula (like CO2 or H2O). If you can still divide them to make them smaller, it is a molecular formula (like C6H12O6, which can be divided by 6).
15.5. Write down one method to find out the molecular mass of a compound without knowing its molecular formula. cc
Answer: We can use a machine called a Mass Spectrometer. It breaks the compound into tiny charged particles and shoots them through a magnet. The machine measures how heavy the particles are and gives us the exact molecular mass directly.
D. Numerical Problems
15.1. Calculate the number of moles, volume and number of molecules in 4.0 g of CH4 at RTP.
Solution:
• Molar mass of CH4 = 12 + (4 × 1) = 16 g/mol.
• Number of Moles = Mass ÷ Molar Mass = 4.0 g ÷ 16 g/mol = 0.25 mol
• Volume at RTP = Moles × 24 dm3 = 0.25 × 24 = 6.0 dm3
• Number of Molecules = Moles × (6.02 × 1023) = 0.25 × 6.02 × 1023 = 1.505 × 1023 molecules
15.2. What mass of NaCl need to be dissolved in 150 cm3 of a solution if the concentration of the solution is 0.4 mol/dm3?
Solution:
• Change volume to dm3: 150 cm3 ÷ 1000 = 0.150 dm3
• Find moles of NaCl needed: Concentration × Volume = 0.4 mol/dm3 × 0.150 dm3 = 0.06 mol
• Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
• Mass needed = Moles × Molar Mass = 0.06 mol × 58.5 g/mol = 3.51 g
15.3. Find out the percentage of sodium in NaHCO3.
Solution:
• Molar mass of NaHCO3 = 23 + 1 + 12 + (3 × 16) = 84 g/mol
• Mass of Sodium (Na) = 23 g
• Percentage of Na = (Mass of Na ÷ Total Molar Mass) × 100 = (23 ÷ 84) × 100 = 27.38%
15.4. A sulphide of iron contains 1.926 g of sulphur and 2.333 g of iron. Find out the empirical formula of this compound.
Solution:
1. Find moles of Iron (Fe): 2.333 g ÷ 55.8 = 0.0418 mol
2. Find moles of Sulphur (S): 1.926 g ÷ 32.1 = 0.0600 mol
3. Divide both by the smallest number of moles (0.0418):
• Fe = 0.0418 ÷ 0.0418 = 1
• S = 0.0600 ÷ 0.0418 = 1.43
4. Multiply both by 2 to make them whole numbers: Fe = 2, S = 3.
• Empirical Formula: Fe2S3
15.5. A compound contains by mass, 40.0% C, 6.71 % H and 53.3 % oxygen. A 0.320 mole of this compound weighs 28.8 g. What is the molecular formula of this compound?
Solution:
Step 1: Find Empirical Formula
• Moles of C = 40.0 ÷ 12 = 3.33 mol
• Moles of H = 6.71 ÷ 1 = 6.71 mol
• Moles of O = 53.3 ÷ 16 = 3.33 mol
• Ratio (divide by 3.33): C = 1, H = 2, O = 1. Empirical Formula = CH2O (Empirical mass = 12+2+16 = 30 g/mol)
Step 2: Find Molecular Formula
• Total Molar Mass = Mass ÷ Moles = 28.8 g ÷ 0.320 mol = 90 g/mol
• Multiplier (n) = Total Molar Mass ÷ Empirical Mass = 90 ÷ 30 = 3
• Molecular Formula = 3 × (CH2O) = C3H6O3
15.6. The formation of ammonia gas is given by the following equation:
3H2 (g) + N2 (g) —> 2NH3 (g)
If you start reacting 12 g of H2 with 64 g of N2, which will be the limiting reactant?
Solution:
• Moles of H2 = 12 g ÷ 2 g/mol = 6 mol
• Moles of N2 = 64 g ÷ 28 g/mol = 2.29 mol
• Check the ratio from the equation to find out moles of NH3:
• For H2:
If 3 moles of hydrogen produce ammonia = 2 moles
Then 1 mole will produce ammonia = 2 ÷ 3 = 0.06 moles
And 6 moles will produce ammonia = 0.06 × 6 = 0.36 moles
• For N2:
If 1 mole of nitrogen produces ammonia = 2 moles
Then 2.29 moles will produce ammonia = 2 × 2.29 = 4.58 moles
• Since H2 (Hydrogen gas) produces less moles of product (ammonia), it is the limiting reactant and it will run out first.
15.7. When 305 g of AgNO3 is reacted with excess of MgCl2 it produces 23.7 g of Mg(NO3)2. What is the percentage yield of the reaction?
2AgNO3 (aq) + MgCl2 (aq) —> Mg(NO3)2 (aq) + 2AgCl (s)
Solution:
• Molar mass of AgNO3 = 170 g/mol
• Moles of AgNO3 used = 305 g ÷ 170 g/mol = 1.79 mol
• The equation shows 2 moles of AgNO3 make 1 mole of Mg(NO3)2. So, expected moles = 1.79 ÷ 2 = 0.895 mol
• Molar mass of Mg(NO3)2 = 148.3 g/mol
• Expected mass (Theoretical Yield) = 0.895 mol × 148.3 g/mol = 132.73 g
• Percentage Yield = (Actual mass ÷ Expected mass) × 100 = (23.7 g ÷ 132.73 g) × 100 = 17.85%
15.8. A sample of metal has a total mass of 3.66 kg and contains 2.45 kg of gold. What is the percentage purity of gold in this sample?
Solution:
• Percentage Purity = (Mass of pure gold ÷ Total mass of sample) × 100
• Percentage Purity = (2.45 kg ÷ 3.66 kg) × 100 = 66.94%
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All Quick Checks
15.1 Calculate the amount of CO2 in 100 dm3 at RTP?
Solution:
• We know that 1 mole of any gas at RTP = 24 dm3.
• Number of Moles = Given Volume ÷ 24 dm3
• Number of Moles = 100 dm3 ÷ 24 dm3 = 4.17 moles
15.2 What mass of CuSO4 is present in 50 cm3 of a 5 x 10–2 mol/dm3 aqueous solution?
Solution:
• Change volume from cm3 to dm3: 50 cm3 ÷ 1000 = 0.05 dm3.
• Concentration = 5 x 10–2 mol/dm3 = 0.05 mol/dm3.
• Find moles of CuSO4: Concentration × Volume = 0.05 mol/dm3 × 0.05 dm3 = 0.0025 moles.
• Molar mass of CuSO4 = 63.5 (Cu) + 32 (S) + (4 × 16) (O) = 159.5 g/mol.
• Mass of CuSO4 = Moles × Molar Mass = 0.0025 mol × 159.5 g/mol = 0.399 g (or approx 0.4 g)
15.3 Calculate the volume of 0.1 mol/dm3 oxalic acid (C2O4H2) solution to exactly neutralize 25 cm3 of 0.1 mol/dm3 KOH solution.
Solution:
• Write the balanced reaction equation: C2O4H2 + 2KOH → K2C2O4 + 2H2O.
• Moles of KOH = Concentration × Volume = 0.1 mol/dm3 × (25 ÷ 1000) dm3 = 0.0025 moles.
• From the equation, 2 moles of KOH need 1 mole of Oxalic Acid. So, moles of oxalic acid needed = 0.0025 ÷ 2 = 0.00125 moles.
• Volume of Oxalic Acid = Moles ÷ Concentration = 0.00125 mol ÷ 0.1 mol/dm3 = 0.0125 dm3.
• Change back to cm3: 0.0125 dm3 × 1000 = 12.5 cm3.
15.4 Calculate the percentage of sulphate ions (SO42-) in Al2(SO4)3.
Solution:
• Total Molar mass of Al2(SO4)3 = (2 × 27) + 3 × [32 + (4 × 16)] = 54 + 3 × [96] = 54 + 288 = 342 g/mol.
• Mass of 3 Sulphate (SO42-) ions = 3 × 96 g = 288 g.
• Percentage of Sulphate = (Mass of Sulphate ions ÷ Total Molar Mass) × 100 = (288 ÷ 342) × 100 = 84.21%
15.5 A 0.5 g sample of compound contains 0.418 g of antimony (Sb) and 0.082 g of oxygen. What is the empirical formula of the compound?
Solution:
1. Find moles of Antimony (Sb) [Atomic Mass = 121.8 g/mol]: 0.418 g ÷ 121.8 = 0.00343 mol.
2. Find moles of Oxygen (O) [Atomic Mass = 16 g/mol]: 0.082 g ÷ 16 = 0.00513 mol.
3. Divide both numbers by the smaller mole value (0.00343):
• Sb = 0.00343 ÷ 0.00343 = 1
• O = 0.00513 ÷ 0.00343 = 1.5
4. Multiply by 2 to make whole numbers: Sb = 2, O = 3.
• Empirical Formula: Sb2O3
15.6 An unknown hydrocarbon contains 85.71% carbon. Its molar mass is 84 g/mol. What is its molecular formula?
Solution:
1. Since it is a hydrocarbon, the rest is Hydrogen: 100% – 85.71% = 14.29% H.
2. Moles of C = 85.71 ÷ 12 = 7.14 mol. Moles of H = 14.29 ÷ 1 = 14.29 mol.
3. Simple ratio (divide by 7.14): C = 1, H = 2. So the Empirical Formula is CH2 (Empirical Mass = 12 + 2 = 14 g/mol).
4. Find the multiplier (n): Total Molar Mass ÷ Empirical Mass = 84 ÷ 14 = 6.
5. Molecular Formula = 6 × (CH2) = C6H12
15.7 Why it is necessary for a chemical equation to be balanced before it can be used in calculation?
Answer: A chemical equation must be balanced to follow the Law of Conservation of Mass, which states that atoms cannot be created or destroyed. A balanced equation gives us the correct recipe (mole ratio) showing exactly how much reactant is needed to form a specific amount of product.
15.8 If you react 100 g of chlorine gas with 500 g of KI, which reactant will be the limiting reactant?
Cl2 (g) + 2KI (aq) –> 2KCl (aq) + I2 (g)
Solution:
• Moles of Cl2 = 100 g ÷ 71 g/mol = 1.41 moles.
• Moles of KI = 500 g ÷ 166 g/mol = 3.01 moles.
• Check the equation ratio:
• For Cl2: 1.41 ÷ 1 = 1.41
• For KI: 3.01 ÷ 2 = 1.505
• Since 1.41 is smaller than 1.505, Cl2 (Chlorine gas) is the limiting reactant because it will run out first.
15.9 Find out the theoretical yield and percent yield of oxygen generated by heating 40 g of KClO3 (M = 122.5). The mass of oxygen gas produced is 14.9g
2KClO3 (s) — > 2KCl (s) + 3O2 (g)
Solution:
• Moles of KClO3 used = 40 g ÷ 122.5 g/mol = 0.327 moles.
• From the equation, 2 moles of KClO3 make 3 moles of O2. So, moles of O2 expected = 0.327 × (3 ÷ 2) = 0.491 moles.
• Molar mass of O2 = 32 g/mol.
• Theoretical Yield = Moles × Molar Mass = 0.491 mol × 32 g/mol = 15.71 g
• Percent Yield = (Actual Mass ÷ Theoretical Yield) × 100 = (14.9 g ÷ 15.71 g) × 100 = 94.84%
