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ICSE Class 10 Chemistry Question 14 of 28

Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 22

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Question

Question 1(2011)

Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane [C3H8].

[C = 12, O = 16, H = 1, Molar Volume = 22.4 dm3 at s.t.p.]

Answer

C3H8+5O2⟶3CO2+4H2O3(12)+8(1)5 vol=44 g5(22.4) lit\begin{matrix} \text{C}_3\text{H}_8 & + &5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 3(12) + 8(1) & & 5 \text{ vol} \\ = 44 \text{ g}& & 5 (22.4) \text{ lit} \\ \end{matrix}

(i) 44 g propane requires 5 x 22.4 lit of oxygen

∴ 8.8 g of propane will require 5×22.444\dfrac{5 \times 22.4}{44} x 8.8 = 22.4 lit.

Hence, 22.4 lit of Oxygen is required.