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Class 10 Chemistry Notes – Chapter 15

Chapter 15: Stoichiometry

Board: Punjab Board

A. Multiple-Choice Questions (MCQs)

1. What is the volume occupied by 15 moles of ammonia gas at RTP?

(a) 224 dm³
(b) 160 dm³
(c) 360 dm³
(d) 265 dm³

Correct Answer: (c) 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

Correct Answer: (a) 0.75 mol

3. What will be the molar concentration of 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

Correct Answer: (a) 0.36 mol/dm³

4. What mass of NaCl is needed to prepare 15 dm³ of a solution with a concentration of 0.7 g/dm³?

(a) 21 g
(b) 10.5 g
(c) 31.5 g
(d) 5.4 g

Correct Answer: (b) 10.5 g

5. Find the concentration of H₂SO₄ in mol/dm³ if 10 cm³ of it reacts with 20 cm³ of NaOH solution having a concentration of 0.1 mol/dm³.

(a) 0.01 mol/dm³
(b) 0.001 mol/dm³
(c) 0.1 mol/dm³
(d) 0.15 mol/dm³

Correct Answer: (c) 0.1 mol/dm³

6. If the molar mass of a compound is 178 g/mol and its empirical formula is C₉H₈O₄, what is its molecular formula?

(a) C₉H₈O₄
(b) C₁₈H₁₆O₈
(c) C₉H₁₄O₄
(d) C₈H₉O₄

Correct Answer: (a) C₉H₈O₄

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

Correct Answer: (d) Natural gas

8. If the actual yield of a compound is 0.198 g and the theoretical yield is 0.217 g, what is the percentage yield?

(a) 50%
(b) 91.2%
(c) 80%
(d) 90%

Correct Answer: (b) 91.2%

B. Short Answer Questions

15.1 What is the molar volume of a gas at RTP?

The molar volume of an ideal gas at room temperature and pressure (RTP) is approximately 24 dm³/mol.

15.2 How is the concept of molar volume useful to us?

Molar volume allows us to convert directly between the volume of a gas and the number of moles without measuring its mass.

15.3 How does molar volume relate to Avogadro’s law?

Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Therefore, one mole of any ideal gas occupies the same molar volume under the same conditions.

15.4 Why may two different compounds have the same empirical formula?

An empirical formula represents the simplest whole-number ratio of atoms in a compound. Different compounds can have the same empirical formula but different molecular formulas. For example, formaldehyde (CH₂O) and glucose (C₆H₁₂O₆) share the empirical formula CH₂O.

15.5 Why is percentage yield important?

Percentage yield measures the efficiency of a chemical reaction by comparing the actual yield with the theoretical yield. It helps determine the success and economic feasibility of chemical processes in industry.

15.6 What is the concentration of a solution in mol/dm³ if it contains 49 g of H₂SO₄ in 1 dm³ of solution?

Given:

Mass of H₂SO₄ = 49 g

Molar mass of H₂SO₄ = 98 g/mol

Volume of solution = 1 dm³

Moles of H₂SO₄:

\[ \text{Moles}=\frac{\text{Mass}}{\text{Molar mass}} \]

\[ =\frac{49}{98}=0.5\text{ mol} \]

Molar concentration:

\[ C=\frac{\text{Moles}}{\text{Volume}} \]

\[ C=\frac{0.5}{1}=0.5\text{ mol/dm}^3 \]

Answer: 0.5 mol/dm³

C. Constructed Response Questions

15.1 How would you identify the limiting reactant in a chemical reaction?

To identify the limiting reactant:

Write the balanced chemical equation.
Calculate the number of moles of each reactant.
Divide the moles of each reactant by its stoichiometric coefficient in the balanced equation.
The reactant with the smallest resulting value is the limiting reactant.

The limiting reactant is consumed first and determines the maximum amount of product that can form.

15.2 Why is it important to determine the purity of a compound used as a medicine?

Determining the purity of a medicinal compound ensures that the medicine contains the correct amount of the active ingredient. It also helps prevent harmful effects caused by impurities and supports the safety and effectiveness of the medicine.

15.3 Will there be a limiting reactant in a reversible reaction?

In a reversible reaction, reactants and products coexist at equilibrium. Unlike a reaction that proceeds to completion, reactants are not necessarily completely consumed. Therefore, the limiting-reactant concept is not applied in the same way as it is for reactions assumed to proceed to completion.

15.4 How do you know whether a formula is an empirical formula or not?

A formula is an empirical formula if its subscripts represent the simplest whole-number ratio of atoms. If all the subscripts can be divided by the same whole number greater than one, the formula can be simplified and is not in empirical form.

For example, C₆H₁₂O₆ simplifies to CH₂O.

15.5 Write one method to determine the molecular mass of a compound without knowing its molecular formula.

One method is Rast’s camphor method, which uses the depression in the freezing point of camphor caused by dissolving a known mass of the compound in it. Other methods use colligative properties, such as elevation of boiling point or depression of freezing point.

D. Numerical Problems

15.1 Calculate the number of moles, volume and number of molecules in 4.0 g of CH₄ at RTP.

Given:

Mass of CH₄ = 4.0 g

Molar mass of CH₄ = 16 g/mol

Molar volume at RTP = 24 dm³/mol

Step 1: Calculate the number of moles.

\[ n=\frac{\text{Mass}}{\text{Molar mass}} \]

\[ n=\frac{4.0}{16}=0.25\text{ mol} \]

Step 2: Calculate the volume at RTP.

\[ V=n\times24 \]

\[ V=0.25\times24=6.0\text{ dm}^3 \]

Step 3: Calculate the number of molecules.

Avogadro’s constant = \(6.022\times10^{23}\) molecules/mol

\[ N=n\times N_A \]

\[ N=0.25\times6.022\times10^{23} \]

\[ N=1.506\times10^{23}\text{ molecules} \]

Answers:

Number of moles = 0.25 mol
Volume at RTP = 6.0 dm³
Number of molecules = \(1.506\times10^{23}\)

15.2 What mass of NaCl must be dissolved in 150 cm³ of solution to obtain a concentration of 0.4 mol/dm³?

Given:

Volume = 150 cm³

Concentration = 0.4 mol/dm³

Molar mass of NaCl = 58.5 g/mol

Step 1: Convert the volume into dm³.

\[ V=\frac{150}{1000}=0.15\text{ dm}^3 \]

Step 2: Calculate the number of moles.

\[ n=C\times V \]

\[ n=0.4\times0.15=0.06\text{ mol} \]

Step 3: Calculate the mass of NaCl.

\[ \text{Mass}=n\times\text{Molar mass} \]

\[ \text{Mass}=0.06\times58.5 \]

\[ \text{Mass}=3.51\text{ g} \]

Answer: 3.51 g of NaCl

15.3 Calculate the percentage of sodium in NaHCO₃.

Molar mass of NaHCO₃:

\[ =23+1+12+3(16) \]

\[ =84\text{ g/mol} \]

Mass of sodium = 23 g/mol

Percentage of sodium:

\[ \%\text{Na}=\frac{\text{Mass of Na}}{\text{Molar mass of NaHCO}_3}\times100 \]

\[ =\frac{23}{84}\times100 \]

\[ =27.38\% \]

Answer: 27.38%

15.4 A sulphide of iron contains 1.926 g of sulphur and 2.333 g of iron. Find its empirical formula.

Given:

Mass of iron = 2.333 g

Mass of sulphur = 1.926 g

Atomic mass of Fe = 55.85 g/mol

Atomic mass of S = 32.1 g/mol

Step 1: Calculate the moles of iron.

\[ n(\text{Fe})=\frac{2.333}{55.85} \]

\[ n(\text{Fe})\approx0.0418\text{ mol} \]

Step 2: Calculate the moles of sulphur.

\[ n(\text{S})=\frac{1.926}{32.1} \]

\[ n(\text{S})=0.0600\text{ mol} \]

Step 3: Find the simplest mole ratio.

Divide both values by the smaller value, 0.0418:

\[ \text{Fe}:\text{S}=\frac{0.0418}{0.0418}:\frac{0.0600}{0.0418} \]

\[ \text{Fe}:\text{S}\approx1:1.435 \]

This ratio is approximately \(1:1.5\). Multiplying both numbers by 2 gives:

\[ \text{Fe}:\text{S}\approx2:3 \]

Empirical formula: Fe₂S₃

15.5 Calculate the molecular formula from mass composition.

Given percentages:

Carbon = 40.0%
Hydrogen = 6.71%
Oxygen = 53.3%

Assume 100 g of the compound.

Step 1: Convert the percentages into masses.

Carbon = 40.0 g
Hydrogen = 6.71 g
Oxygen = 53.3 g

Step 2: Calculate the moles of each element.

\[ n(\text{C})=\frac{40.0}{12}\approx3.33 \]

\[ n(\text{H})=\frac{6.71}{1.008}\approx6.66 \]

\[ n(\text{O})=\frac{53.3}{16}\approx3.33 \]

Step 3: Find the simplest mole ratio.

Divide each value by the smallest value, approximately 3.33:

\[ \text{C}:\text{H}:\text{O}\approx1:2:1 \]

Therefore, the empirical formula is:

\[ \boxed{\text{CH}_2\text{O}} \]

Empirical formula mass:

\[ 12+2(1)+16=30\text{ g/mol} \]

Note: The supplied text does not clearly provide a reliable molecular mass for this question. Therefore, the molecular formula cannot be determined conclusively from the provided information alone.

The molecular formula is calculated using:

\[ n=\frac{\text{Molecular mass}}{\text{Empirical formula mass}} \]

\[ \text{Molecular formula}=(\text{CH}_2\text{O})_n \]

15.6 Limiting Reactant Problem

Nitrogen reacts with hydrogen to form ammonia according to the balanced equation:

\[ \text{N}_2+3\text{H}_2\rightarrow2\text{NH}_3 \]

Given:

Mass of H₂ = 12 g

Mass of N₂ = 64 g

Molar mass of H₂ = 2 g/mol

Molar mass of N₂ = 28 g/mol

Step 1: Calculate the moles of hydrogen.

\[ n(\text{H}_2)=\frac{12}{2}=6\text{ mol} \]

Step 2: Calculate the moles of nitrogen.

\[ n(\text{N}_2)=\frac{64}{28}\approx2.286\text{ mol} \]

Step 3: Determine the required mole ratio.

According to the balanced equation, 1 mole of N₂ reacts with 3 moles of H₂.

For 6 moles of H₂, the required nitrogen is:

\[ n(\text{N}_2)=\frac{6}{3}=2\text{ mol} \]

Available nitrogen = 2.286 mol

Required nitrogen = 2 mol

Since sufficient nitrogen is available to react with all the hydrogen, hydrogen is consumed first.

Answer: H₂ is the limiting reactant, and N₂ is the excess reactant.

15.7 Percentage Yield Calculation

Given:

Actual yield = 23.7 g

Theoretical yield must be calculated from the balanced chemical equation and the quantities of reactants provided in the original question.

Percentage yield is calculated using:

\[ \%\text{Yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100 \]

Substituting the known actual yield:

\[ \%\text{Yield}=\frac{23.7}{\text{Theoretical yield}}\times100 \]

Note: The reactant quantities and balanced equation for this problem were not included in the supplied text, so the numerical percentage yield cannot be calculated reliably.

15.8 Calculate the percentage purity of gold.

Given:

Mass of pure gold = 2.45 kg

Total mass of sample = 3.66 kg

Formula:

\[ \%\text{Purity}=\frac{\text{Mass of pure substance}}{\text{Total mass of sample}}\times100 \]

Substituting the values:

\[ \%\text{Purity}=\frac{2.45}{3.66}\times100 \]

\[ \%\text{Purity}\approx66.94\% \]

Answer: 66.94%

Notes by Sir Dilawar

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