# Calcium hydride combines with water according to the equation: CaH_2(s) + 2H_2O(ℓ) -> 2H_2(g) + Ca(OH)_2(s)? Beginning with with 84.0 g of CaH_2(s) and 36.0 g of H_2O, what volume of H_2 will be produced at 273 K and a pressure of 1520 torr?

Apr 24, 2016

$22.5 L$

#### Explanation:

Calculate the number of moles of each of the reactants to determine what the limiting reagent is

${M}_{r} \left({H}_{2} O\right) = 18.0 \frac{g}{m o l}$

n(H_2O)=(36.0g)/(18.0g·mol^-1)=2molH_2O

${M}_{r} \left(C a {H}_{2}\right) = 42 \frac{g}{m o l}$

n(H_2O)=(84.0g)/(42g·mol^-1)=2molCaH_2

You have two moles of each, but in order to fully react you would need four moles of ${H}_{2} O$, since there is a $\frac{2}{1}$ ratio. This means that ${H}_{2} O$ is the limiting reagent, and for molar ratios we should use ${H}_{2} O$ rather than the excess $C a {H}_{2}$.

There is a $\frac{2}{2}$ ratio of water to hydrogen gas, which is the same as $\frac{1}{1}$ - for every mole of water, one mole of hydrogen is produced. For two moles of water reacted in the question, two moles of hydrogen gas are produced.

Now you can use the ideal gas equation, inserting all the values we know so far

$P V = n R T$

where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the constant to convert between units, and $T$ is the temperature.

We know that $n = 2 m o l$, $P = 1520 \text{torr}$ and $T = 273 K$. Since we also know the units of these values, we can find $R$ from a table as $62.364 \frac{L \cdot \text{Torr}}{m o l \cdot K}$.

Since the question is asking the volume, we can rearrange the ideal gas equation and plug in these values to get

$V = \frac{n R T}{P}$

$= \frac{2 \cdot 62.364 \cdot 273}{1520} \approx 22.4 L$