calcium carbide reacts with water to form ethyne. What mass of calcium carbide (grams) would react to give \(1.2 \mathrm{dm}^3\) of ethyne at STP? \([\mathrm{Ca}=40, \mathrm{C}=12, \mathrm{H}=1]\)
A. \(60 \mathrm{~g}\) B. \(64 \mathrm{~g}\) C. \(32 \mathrm{~g}\) D. \(3.2 \mathrm{~g}\)
Correct Answer: D
Explanation
\(\mathrm{CaC}_2+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CaO}+\mathrm{C}_2 \mathrm{H}_2\) I mole of calcium carbide produces I mole of ethyne. \(\mathrm{CaC}_2=40+(12 \times 2)=64 \mathrm{gmol}^{-1}\). Then, \(64 \mathrm{~g}_{\text {of }} \mathrm{CaC}_2\) ( 1 mole) produces \(22.4 \mathrm{dm}^3\) of \(\mathrm{C}_2 \mathrm{H}_2\) (1 mole) at stp. \(1.12 \mathrm{dm}^3\) of \(\mathrm{C}_2 \mathrm{H}_2\) will require \(\frac{64 \mathrm{~g}}{22.4 \mathrm{dm}^3} \times 1.12 \mathrm{dm}^3=3.2 \mathrm{~g}\)
