Chapter 24, Problem 15QP

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction of ${\text{CaH}}_2$

and ${\text{H}}_{\text{2}}\text{O}$

is to be shown and mass of ${\text{CaH}}_2$

required to produce the given amount of ${\text{H}}_2$

is to be calculated.

Concept introduction:

The balanced chemical equation must contain an equal number of similar atoms on each side.

The Ideal gas equation is given as:

$PV=n\text{R}T$

Here, $P,V,n,\text{R,}\text{and}T$

are pressure, volume, number of moles, universal gas constant, and temperature, respectively.

Answer to Problem 15QP

Solution: The reaction is as follows:

${\text{CaH}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}\to {\text{Ca(OH)}}_2+2{\text{H}}_2$

$22.7\text{g}{\text{CaH}}_2$

Explanation of Solution

Given information:

Temperature ${20}^0\text{C}$

=

Pressure $\text{746}\text{mmHg}$

Volume of gas $\text{26}\text{.6 L}$

Convert temperature Celsius to Kelvin

${20}^0\text{C}=20+273=293\text{K}$

The reaction of ${\text{CaH}}_2$

and ${\text{H}}_{\text{2}}\text{O}$

produces ${\text{Ca(OH)}}_2$

and hydrogen gas.

The balanced chemical equation of ${\text{CaH}}_2$

and ${\text{H}}_{\text{2}}\text{O}$

is written as follows:

${\text{CaH}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}\to {\text{Ca(OH)}}_2+2{\text{H}}_2$

Hence, the balanced equation is stated.

The Ideal gas equation is given as:

$PV=n\text{R}T$

Rearrange above expression and substitute the value $0.0821{\text{L atm mol}}^{-1}{\text{ K}}^{-1}$ for R, $293\text{K}$

for $T$, $746\text{mmHg}$

for $P$ and $26.4\text{L}$

for $V$ in the expression as:

$n=\frac{PV}{\text{R}T}$

$\begin{array}{c}n=\left(\frac{746\cancel{\text{mmHg}}\times \frac{1\cancel{\text{atm}}}{\cancel{760\text{mmHg}}}\times 26.4\cancel{\text{L}}}{0.0821\cancel{\text{L}}\cancel{\text{atm }}{\text{mol}}^{-1}\cancel{{\text{K}}^{-1}}\times 293\cancel{\text{K}}}\right)\\ =1.08\text{mol}{\text{H}}_2\end{array}$

The molar mass of ${\text{CaH}}_2=42.10\text{g}$.

The mass of ${\text{CaH}}_2$

required in grams to produce $1.08\text{mol}{\text{H}}_2$

by the mole ratio from the balanced chemical equation is calculated as follows:

$\begin{array}{c}{m}_{{\text{CaH}}_{\text{2}}}=\left(1.08\cancel{\text{mol}{\text{H}}_2}\right)\left(\frac{\cancel{1\text{mol}{\text{CaH}}_2}}{2\cancel{\text{mol}{\text{H}}_2}}\right)\left(\frac{42.10\text{g}{\text{CaH}}_2}{\cancel{1\text{mol}{\text{CaH}}_2}}\right)\\ =22.7\text{g}\end{array}$

Conclusion

The balanced chemical equation for the reaction of ${\text{CaH}}_2$

and ${\text{H}}_{\text{2}}\text{O}$

is written as required, and the mass of ${\text{CaH}}_2$

required to produce the given amount of ${\text{H}}_2$ is $22.7\text{ g }$.