# What mass of calcium hydride is obtained from a 5.0*g mass of calcium metal?

Mar 16, 2017

Approx. $5 \cdot g$................

#### Explanation:

Calcium hydride is formed by direct combination of its elements at elevated temperature:

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

Given that the dihydrogen was in excess, moles of calcium hydride is equivalent to moles of calcium:

$\text{Moles of calcium} = \frac{5.0 \cdot g}{40.08 \cdot g \cdot m o {l}^{-} 1} = 0.125 \cdot m o l$

Given the $1 : 1$ stoichiometry, we form $0.125 \cdot m o l$ $C a {H}_{2}$, and this represents a mass of $0.125 \cdot \cancel{m o l} \times 42.10 \cdot g \cdot \cancel{m o {l}^{-} 1} = 5.25 \cdot g$.

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