ICSE Mole Concept and Stoichiometry Numericals for Class 10 Chemistry

ICSE Mole Concept and Stoichiometry Numericals for Class 10 Chemistry

In this article, we are providing Important Numericals based on Mole Concept and Stoichiometry for ICSE Class 10 Chemistry.

In chemistry, the mole is a fundamental unit that is used to measure the amount of substance. This quantity is also referred as the chemical amount. The word ‘stoichiometry’ is derived from the combination of two Greek words i.e., stoicheion (means element) and metron (means measure).

Stoichiometry, thus deals with the calculations of masses (sometimes volume) of the reactants and the products involved in a chemical reaction. In this chapter, we will deal with these chemical calculations using the mole concept and stoichiometry to understand molecules and chemical reactions involved.

Q1. Calculate the volume occupied by \(3.4 \, \text{g}\) of ammonia at S.T.P.
Sol:
Given:
Gram molecules of \(\mathrm{NH}_3 = (1 \times 14) + (3 \times 1) = 14 + 3 = 17 \, \text{g}\)
Mass of one mole of ammonia \(= 17 \, \text{g}\)
Molar volume \(= 22.4 \, \text{litre}\)
Thus, 17 grams of \(\mathrm{NH}_3\) occupy \(22.4 \, \text{litre}\).
\[ \text{Volume occupied by } 3.4 \, \text{gram} = \frac{3.4 \times 22.4}{17} = 4.48 \, \text{litre} \] Therefore, the volume occupied by \(3.4 \, \text{g}\) of \(\mathrm{NH}_3\) at S.T.P. is \(4.48 \, \text{litre}\).

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Q2. What weight of zinc is needed to produce \(100 \, \text{ml}\) of dry hydrogen at S.T.P. from dilute sulphuric acid solution?
Sol:
The chemical equation representing the reaction is:
\[ \underset{(65 \, \text{g})}{\mathrm{Zn} + \mathrm{H}_2 \mathrm{SO}_4} \rightarrow \mathrm{ZnSO}_4 + \underset{(22400 \, \text{ml})}{\mathrm{H}_2} \] Since \(22400 \, \text{ml}\) of hydrogen is liberated at S.T.P. from \(65 \, \text{g}\) of zinc, therefore \(100 \, \text{ml}\) of hydrogen is liberated at S.T.P. from:
\[ = \frac{65}{22400} \times 100 = 0.29 \, \text{g} \] Hence, 0.29 g of zinc is needed to produce \(100 \, \text{ml}\) of dry hydrogen at S.T.P. from dilute sulphuric acid solution.

Q3. What weight and volume of oxygen at S.T.P. will be given when \(18 \, \text{g}\) of water is electrolyzed?
Sol:
The decomposition of water can be represented as:
\[ \begin{aligned} & 2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_2 + \mathrm{O}_2 \\ & (36 \, \text{g}) \quad \quad \quad \quad (4 \, \text{g}) \quad \quad \quad \quad (32 \, \text{g}) \end{aligned} \] Since \(36 \, \text{g}\) of water yields \(32 \, \text{g}\) of oxygen:
\[ \text{18 g of water will yield} = \frac{32}{36} \times 18 = 16 \, \text{g} \text{ of oxygen} \] Again, \(32 \, \text{g}\) of oxygen occupies \(22.4 \, \text{litre}\) volume at S.T.P.
\[ \text{16 g of oxygen will occupy} = \frac{22.4}{32} \times 16 = 11.2 \, \text{litre} \] Hence, when \(18 \, \text{g}\) of water is electrolyzed, it produces \(16 \, \text{g}\) of oxygen, and the volume of this oxygen at S.T.P. is \(11.2 \, \text{litre}\).

Q4. (i) How many grams of sulfuric acid will be required to completely neutralize \(16.0 \, \text{g}\) of caustic soda?
Sol:
The complete reaction between \(\mathrm{H}_2 \mathrm{SO}_4\) and \(\mathrm{NaOH}\) is represented by:
\[ \begin{aligned} 2 \mathrm{NaOH} + \mathrm{H}_2 \mathrm{SO}_4 &\rightarrow \mathrm{Na}_2 \mathrm{SO}_4 + 2 \mathrm{H}_2 \\ 2[23 + 16 + 1] \, \text{g of} \, \mathrm{NaOH} &\text{ react with} \, [2 \times 1 + 32 + 4 \times 16] \, \text{g of} \, \mathrm{H}_2 \mathrm{SO}_4 \end{aligned} \] or \(80 \, \text{g}\) of \(\mathrm{NaOH}\) react with \(98 \, \text{g}\) of \(\mathrm{H}_2 \mathrm{SO}_4\). \[ 16 \, \text{g of} \, \mathrm{NaOH} \, \text{react with} \, \frac{98 \times 16}{80} = 19.6 \, \text{g} \, \text{of} \, \mathrm{H}_2 \mathrm{SO}_4 \] (ii) What volume of \(\mathrm{H}_2\) be evolved at S.T.P.?
\(80 \, \text{g}\) of \(\mathrm{NaOH}\) formed at S.T.P. \(= 2 \times 22.4 \, \text{litre}\) of \(\mathrm{H}_2\). \[ 16 \, \text{g of} \, \mathrm{NaOH} \, \text{formed at S.T.P.} = \frac{2 \times 22.4 \times 16}{80} = 8.96 \, \text{litre} \, \text{of} \, \mathrm{H}_2 \] Hence, (i) \(19.6 \, \text{g}\) of \(\mathrm{H}_2 \mathrm{SO}_4\) is required to completely neutralize \(16.0 \, \text{g}\) of \(\mathrm{NaOH}\), and (ii) \(8.96 \, \text{litre}\) of \(\mathrm{H}_2\) will be evolved at S.T.P.

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Q5. (i) Iron pyrites has formula \(\mathrm{FeS}_2\). What mass of sulfur is contained in \(30 \, \text{g}\) iron pyrites?
Sol:
\[ \begin{aligned} \text{Gram molecular mass of } \mathrm{FeS}_2 &= 56 + 2 \times 32 = 120 \, \text{g} \\ \text{Mass of sulfur} &= 2 \times 32 = 64 \, \text{g} \end{aligned} \] Sulfur contained in \(120 \, \text{g}\) of \(\mathrm{FeS}_2\) is \(64 \, \text{g}\). \[ \text{Sulfur contained in } 30 \, \text{g} \text{ of } \mathrm{FeS}_2 = \frac{64}{120} \times 30 = 16 \, \text{g} \] (ii) When roasted, iron pyrites gives sulfur dioxide according to the equation: \[ 4 \mathrm{FeS}_2 + 11 \mathrm{O}_2 \rightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3 + 8 \mathrm{SO}_2 \] Now, 4 moles of \(\mathrm{FeS}_2\) give 8 moles of \(\mathrm{SO}_2\) or 1 mole of \(\mathrm{FeS}_2\) gives 2 moles of \(\mathrm{SO}_2\).
\[ \text{Molecular mass of } \mathrm{FeS}_2 = 56 + 32 \times 2 = 120 \] Thus, \(120 \, \text{g}\) of \(\mathrm{FeS}_2\) gives \(2 \times 22.4\) litres of \(\mathrm{SO}_2\) at S.T.P.
\[ \text{Volume of } \mathrm{SO}_2 \text{ obtained from } 120 \, \text{g} \text{ of } \mathrm{FeS}_2 = 2 \times 22.4 \, \text{litres} \, \text{at S.T.P.} \] \[ \text{Volume of } \mathrm{SO}_2 \text{ obtained from } 30 \, \text{g} \text{ of } \mathrm{FeS}_2 = \frac{2 \times 22.4}{120} \times 30 = 11.2 \, \text{litres} \, \text{at S.T.P.} \] Hence, (i) \(16 \, \text{g}\) of sulfur is contained in \(30 \, \text{g}\) of iron pyrites, and (ii) \(11.2 \, \text{litres}\) of \(\mathrm{SO}_2\) at S.T.P. would be liberated by roasting \(30 \, \text{g}\) of iron pyrites.

Download 65 Numericals Based on Mole Concept and Stoichiometry

Numericals on All Chapter Topics Covered

Here is the list of topics that we have covered in this article.

• Gay-Lussac’s Law
• Avogadro’s Law
• Mole Concept
• Atomicity
• Percentage Composition
• Empirical Formula of a
• Compound
• Molecular Formula of a
• Compound
• Stoichiometry and
• Stoichiometric Calculations

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