What mass of \(Ca{H_2}\) is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 °C and 0.8 atm pressure with a volume of 4.5 L? The balanced equation is:

\(Ca{H_2}\left( s \right) + 2{H_2}O\left( l \right) \to Ca{\left( {OH} \right)_2}\left( {aq} \right) + 2{H_2}\left( g \right)\)

A salt like compound named calcium hydride is white and crystalline when pure but is usually obtained in gray to gray-brown lumps.

It is used chiefly as a reducing agent in the preparation of powdered metals as a portable source of hydrogen, and as a drying agent.

The ideal gas law is\(P \times V = n \times R \times T.\)

Here,

• \(P\)is the pressure,
• \(V\)is the volume,
• \(R\)is the constant,and
• \(T\)is the temperature.
• The number of moles of \({{\rm{H}}_2}\) is labeled as n.

Replace the known values.

\(\begin{aligned}{}n &= \frac{{PV}}{{RT}}\\ &= \frac{{0.8{\rm{ atm }} \times 4.5L}}{{0.0821\frac{{{\rm{ L}} \times {\rm{atm }}}}{{{\rm{ mol}} \times {\rm{K }}}} \times 293K}}\\ &= \frac{{3.6atm{\rm{ }}L}}{{24.0553\frac{{L \times atm}}{{{\rm{ mol }}}}}}\\ &= 0.1496{\rm{mol }}{{\rm{H}}_2}.\end{aligned}\)

If one mol of \({\rm{Ca}}{{\rm{H}}_2}\) gives 2 moles of\({{\rm{H}}_2}\), then the number of moles of \({\rm{Ca}}{{\rm{H}}_2}\)is evaluated as:

\(\begin{aligned}{}n = \frac{{0.1496}}{2}\\\;\; &= 0.0748\;{\rm{moles of Ca}}{{\rm{H}}_{\rm{2}}}.\end{aligned}\)

The formula for calculating mass is:

\({\rm{Mass = Molar mass \times Number of moles}}\)

\(\begin{aligned}{}Mass {\rm{ }} of {\rm{ Ca}} {{\rm{H}}_2} & = 0.0748\;{\rm{mol}} \times 42.094\frac{{\rm{g}}}{{{\rm{mol}}}}\\ &= 3.149\;\;{\rm{g}}{\rm{.}}\end{aligned}\)

Therefore, the mass of\(Ca{H_2}\)is 3.149g.