# Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. How many grams of calcium hydride are needed to form 8.5 grams of hydrogen?

##### 1 Answer
Mar 20, 2016

Approx. $85$ $g$ of calcium hydride are required

#### Explanation:

$C a {H}_{2} \left(s\right) + 2 {H}_{2} O \left(l\right) \rightarrow C a {\left(O H\right)}_{2} \left(a q\right) + 2 {H}_{2} \left(g\right) \uparrow$

The balanced equation gives a 1:1 stoichimetry between calcium hydride and dihydrogen gas, We need to (i) work out the molar quantity of dihydrogen, and (ii) convert this into an equivalent mas of $C a {H}_{2}$.

$\text{Moles of } {H}_{2} \left(g\right)$ $=$ $\frac{8.5 \cdot g}{2.0 \cdot g \cdot m o {l}^{-} 1}$ $\approx$ $4$ $m o l$.

So, if 4 mol of calcium hydride are required. This gives a mass of $2 \cdot m o l \times 42.09 \cdot g \cdot m o {l}^{-} 1 \text{ calcium hydride}$ $=$ ??g