What mass, in grams, of hydrogen gas forms during the complete reaction of \({\bf{10}}.{\bf{01}}{\rm{ }}{\bf{g}}\) calcium with water?

Calcium hydride\(\left( {{\bf{Ca}}{{\bf{H}}_{\bf{2}}}} \right)\)vigorously reacts with water, releasing hydrogen gas.Controlling the rate of hydrogen production is an absolute requirement for the development of an effective hydrogen storage system.

The number of moles of calcium is evaluated as:

\(\begin{array}{l}n(Ca) = \frac{{m(NaCl)}}{{M(NaCl)}}\\n(Ca) = \frac{{10.01g}}{{40.078g/mol}}\;\;\\n(Ca) = 0.25\;{\rm{mol}}\end{array}\)

The reaction:

\({\rm{Ca}}(s) + 2{{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {\rm{Ca}}{({\rm{OH}})_2}(aq) + {{\rm{H}}_2}(g)\)

To calculate the number of moles produced by\({{\rm{H}}_2}\),multiply the numberof moles of\({\rm{Ca}}\)by the stoichiometric ratio.

\(\frac{{\nu (Ca)}}{{\nu \left( {{H_2}} \right)}} = \frac{2}{1}\;\;\;\)

\(\begin{array}{l}n\left( {{H_2}} \right) = n({\rm{Ca}}) \cdot \frac{{\nu \left( {{{\rm{H}}_2}} \right)}}{{\nu ({\rm{Ca}})}}\;\\n\left( {{H_2}} \right) = 0.25mol \cdot \frac{1}{2}\;\;\\n\left( {{H_2}} \right) = 0.125\;{\rm{mol}}\end{array}\)

Finally, calculate the mass of\({{\rm{H}}_2}\)in the following way:

\(\begin{array}{}m\left( {{H_2}} \right) &= {\rm{Number of moles \times Molar mass}}\\m\left( {{H_2}} \right) &= 0.125\;{\rm{mol}} \times 2.016\;{\rm{g}}/{\rm{mol}}\\m\left( {{H_2}} \right)& = 0.252g\end{array}\)