Question

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

Short answer

The mass of\(m\left( {{H_2}} \right) = 0.252g\).

Step-by-step solution

Reaction of calcium hydride 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.

Moles of calcium

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:

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

Moles of hydrogen

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

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

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

Mass of hydrogen

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

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