# A certain mass of CaC_2 reacts completely with water to give 64.5 L of C_2H_2 at 50.0 C and 1.00 atm. If the same mass of CaC_2 reacts completely at 400.0 C and P = 1.00 atm, what volume of C_2H_2 will be collected at the higher temperature?

## Calcium carbide ($C a {C}_{2}$) reacts with water to produce acetylene (${C}_{2} {H}_{2}$) according to the equation $C a {C}_{2} + 2 {H}_{2} O \to C a {\left(O H\right)}_{2} + {C}_{2} {H}_{2}$

Feb 14, 2017

The volume of acetylene is 134 L.

#### Explanation:

This is really a disguised Charles' Law problem.

In each case, we have the same mass and therefore the same number of moles of calcium carbide.

Hence, we will have the same number of moles of acetylene in each case.

Thus, $n$ and $P$ are constant; the only variables are $V$ and $T$.

Charles' Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this to get

V_2 = V_1 × T_2/T_1

In this problem,

${V}_{1} = \text{64.5 L"; T_1 = color(white)(ll)"50.0 °C" = "323.15 K}$
V_2 = ?color(white)(mml); T_2 = "400.0 °C" = "673.15 K"

${V}_{2} = \text{64.5 L" × (673.15 color(red)(cancel(color(black)("K"))))/(323.15 color(red)(cancel(color(black)("K")))) = "134 L}$