# Question #1bbbc

Jan 17, 2017

You will require a sample of impure $C a {C}_{2}$ that is 10.02 g in mass.

#### Explanation:

The equation for acetylene production is

$C a {C}_{2} + 2 {H}_{2} O \rightarrow {C}_{2} {H}_{2} + C a {\left(O H\right)}_{2}$

At STP, 3.36 L of any gas represents $3.36 \div 22.4 = 0.15 m o l$

Since the ratio between $C a {C}_{2}$ and ${C}_{2} {H}_{2}$ is 1 to 1 in the equation, we require 0.15 mol of calcium carbide to do the job.

The molar mass of $C a {C}_{2}$ is 64.1 g, so 0.15 mol would have a mass of 9.615 g.

Since the sample is 4% impure, the 9.615 g must represent 96% of the mass of the impure sample, so the sample must be

$9.615 \div 0.96 = 10.02 g$ in size.