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JEE Chemistry 2025 Question with Solution

When 81.0 g81.0 \, \text{g} of aluminium is allowed to react with 128.0 g128.0 \, \text{g} of oxygen gas, the mass of aluminium oxide produced in grams is _____ (nearest integer).

Answer

Correct answer:153

Step-by-step solution

Standard Method

Given: Mass of aluminium = 81.0 g81.0 \, \text{g}, mass of oxygen gas = 128.0 g128.0 \, \text{g}.

Find: Mass of aluminium oxide produced.

The balanced chemical equation is

4Al+3O2→2Al2O34\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3

Molar masses:

  • Al=27 g/mol\text{Al} = 27 \, \text{g/mol}
  • O2=32 g/mol\text{O}_2 = 32 \, \text{g/mol}
  • Al2O3=2×27+3×16=102 g/mol\text{Al}_2\text{O}_3 = 2\times 27 + 3\times 16 = 102 \, \text{g/mol}

Moles present:

n(Al)=81.027=3.0 moln(\text{Al}) = \frac{81.0}{27} = 3.0 \, \text{mol} n(O2)=128.032=4.0 moln(\text{O}_2) = \frac{128.0}{32} = 4.0 \, \text{mol}

From the reaction, 44 moles of Al\text{Al} require 33 moles of O2\text{O}_2. Therefore, 3.03.0 moles of Al\text{Al} require

3.0×34=2.25 mol of O23.0 \times \frac{3}{4} = 2.25 \, \text{mol of } \text{O}_2

Since available oxygen is 4.0 mol4.0 \, \text{mol}, oxygen is in excess and aluminium is the limiting reagent.

From the stoichiometric ratio,

n(Al2O3)=3.0×24=1.5 moln(\text{Al}_2\text{O}_3) = 3.0 \times \frac{2}{4} = 1.5 \, \text{mol}

Hence, mass of aluminium oxide formed is

1.5×102=153.0 g1.5 \times 102 = 153.0 \, \text{g}

Therefore, the mass of aluminium oxide produced is 153 g153 \, \text{g}. Hence the required numerical answer is 153.

Use stoichiometry to calculate the mass of the product based on the limiting reagent in the reaction.

Alternative Working

Given: 81.0 g81.0 \, \text{g} of aluminium and 128.0 g128.0 \, \text{g} of oxygen gas.

Find: Mass of Al2O3\text{Al}_2\text{O}_3 formed.

The reaction is

4Al+3O2→2Al2O34\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3

Moles of aluminium:

81.027.0=3\frac{81.0}{27.0} = 3

Moles of oxygen:

128.032.0=4\frac{128.0}{32.0} = 4

According to stoichiometry, 44 moles of aluminium produce 22 moles of aluminium oxide. So 33 moles of aluminium produce

24×3=1.5 moles of Al2O3\frac{2}{4} \times 3 = 1.5 \text{ moles of } \text{Al}_2\text{O}_3

Molar mass of aluminium oxide:

2(27.0)+3(16.0)=102.0 g/mol2(27.0) + 3(16.0) = 102.0 \, \text{g/mol}

Therefore, the mass formed is

1.5×102.0=153 g1.5 \times 102.0 = 153 \, \text{g}

Thus, the mass of aluminium oxide produced is 153 g153 \, \text{g}.

Common mistakes

  • Treating oxygen as the limiting reagent without checking mole ratio is incorrect because the reaction must be compared using the balanced equation. First convert both reactants to moles and then identify the limiting reagent from stoichiometric requirement.

  • Using the molar mass of atomic oxygen as 16 g/mol16 \, \text{g/mol} for oxygen gas is wrong because the reactant is O2\text{O}_2, whose molar mass is 32 g/mol32 \, \text{g/mol}. Always use the molar mass of the species actually present in the equation.

  • Calculating the product moles as equal to aluminium moles is incorrect because the coefficient ratio is not 1:11:1. From 4Al→2Al2O34\text{Al} \rightarrow 2\text{Al}_2\text{O}_3, product moles are half the aluminium moles.

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