# Question #d97b5

##### 1 Answer
Jan 7, 2017

The balanced equation of the reaction for the preparation of ethyne (${C}_{2} {H}_{2}$) by the action of water on $C a {C}_{2}$ is as follows.

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

As per above equation 1 mol ${C}_{2} {H}_{2}$ is produced from 1 mol $C a {C}_{2}$.

Molar mass of $C a {C}_{2} = \left(40 + 2 \cdot 12\right) = 64 \frac{g}{\text{mol}}$

The given amount of $C a {C}_{2} = 20 g = \frac{20}{64} m o l = \frac{5}{16} m o l$.

So the amount of ethyne ${C}_{2} {H}_{2}$ produced $= \frac{5}{16} m o l$.
The molar mass of ethyne ${C}_{2} {H}_{2} = 2 \times 12 + 2 \times 1 = 26 \frac{g}{\text{mol}}$.

So the mass of ethyne ${C}_{2} {H}_{2}$ produced

$= \frac{5}{16} m o l \times 26 \frac{g}{\text{mol}} = 8.125 g$.