Q19E - Chemistry (OpenStax)

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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:

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

 

Step 3: Moles of hydrogen

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

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

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

Step 4: Mass of hydrogen

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

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