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Chemistry

From the equation:

C + 2H2SO4 ⟶ CO2 + 2H2O + 2SO2

Calculate:

(i) The mass of carbon oxidized by 49 g of sulphuric acid.

(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.

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Answer

C+2H2SO4⟶CO2+2H2O+2SO212g2[2(1)+32+4(16)]2[2+32+64]196g\begin{matrix} \text{C} & + &2\text{H}2\text{SO}4 & \longrightarrow & \text{CO}2 & + & 2\text{H}2\text{O}& + & 2\text{SO}_2 \ 12 \text{g} && 2[2(1) + 32 + 4(16)] \ && 2[2 + 32 + 64] \ && 196 \text{g} \ \end{matrix}

(i) 196 g of sulphuric acid oxidizes 12 g carbon

∴ 49 g of sulphuric acid will oxidize = 12196\dfrac{12}{196} x 49 = 3 g

Hence, 3 g of carbon is oxidized.

(ii) 12 g carbon liberates 2 vol = (2 x 22.4) lit of SO2

∴ 3 g of carbon will liberate 2×22.412\dfrac{2 \times 22.4}{12} x 3 = 11.2 lit of SO2.

Hence, 11.2 lit of SO2 is liberated.

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