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?

Question

Calcium carbide ( $CaC_2$ ) reacts with water to produce acetylene ( $C_2H_2$ ) according to the equation
$CaC_2 + 2H_2O -> Ca(OH)_2 + C_2H_2$

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Answer 1 · 2017-02-14T22:12:29

The volume of acetylene is 134 L.

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

$color(blue)(bar(ul(|color(white)(a/a) V_1/T_1 = V_2/T_2color(white)(a/a)|)))" "$

We can rearrange this to get

$V_2 = V_1 × T_2/T_1$

In this problem,

$V_1 = "64.5 L"; T_1 = color(white)(ll)"50.0 °C" = "323.15 K"$
$V_2 = ?color(white)(mml); T_2 = "400.0 °C" = "673.15 K"$

∴ $V_2 = "64.5 L" × (673.15 color(red)(cancel(color(black)("K"))))/(323.15 color(red)(cancel(color(black)("K")))) = "134 L"$