Chapter 22, Problem 22.15QP

Interpretation Introduction

Interpretation:

Balanced equation for the reaction of ${\text{CaH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\text{O}$ and grams ${\text{CaH}}_{\text{2}}$ of needed for the formation of 26.4L of ${\text{H}}_{\text{2}}$ has to be determined at ${\text{20}}^{\text{o}}\text{C}$ and $\text{746}\text{mm}\text{}\text{Hg}$ .

Concept introduction:

Balanced chemical equation

The mass of products should be equal to mass of its starting materials this is governed by law of conservation of mass.

$\begin{array}{l}\text{Mass}\text{of}\text{starting}\text{materials}\left(\text{Reactants}\right)\text{=}\text{Mass}\text{of}\text{products}\\ {\text{N}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\to {\text{NH}}_{\text{3}}\end{array}$

In the above reaction 2 nitrogen atoms are there in reactant side but 1 nitrogen in product side, likewise 2 hydrogen’s in reactant side but 3 hydrogen’s in product side so this is unbalanced equation by putting coefficients for each elements in both sides we can balance this reaction as follows:

${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to {\text{2NH}}_{\text{3}}$

Answer to Problem 22.15QP

• Balanced equation for the reaction of ${\text{CaH}}_{\text{2}}$ with ${\text{H}}_{\text{2}}\text{O}$ is given below:${\text{CaH}}_{\text{2}}{}_{\left(\text{s}\right)}\text{+}{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}\text{+}{\text{2H}}_{\text{2}}{}_{\left(\text{g}\right)}$
• The amount of ${\text{CaH}}_{\text{2}}$ needed to produce 26.4L of ${\text{H}}_{\text{2}}$ is 22.7 g

Explanation of Solution

The balanced equation for the given starting materials is given below:

${\text{CaH}}_{\text{2}}{}_{\left(\text{s}\right)}\text{+}{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}\text{+}{\text{2H}}_{\text{2}}{}_{\left(\text{g}\right)}$

The reaction of calcium hydride with water gives calcium hydroxide and hydrogen gas as products. By putting coefficients of the starting materials and products the mass of starting materials will be equal to the mass of the products.

Pressure given as $\text{746}\text{mm}\text{}\text{Hg}$

Temperature given as ${\text{20}}^{\text{o}}\text{C}$

Gas constant value is $\text{0}\text{.0821}\raisebox{1ex}{$\text{L}\cdot \text{atm}$}\!\left/ \!\raisebox{-1ex}{$\text{K}\cdot \text{mol}$}\right.$

Calculate the number of moles of ${\text{H}}_{\text{2}}$

Ideal gas equation is given as follows,

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{P}\text{-}\text{pressure of gas; V}\text{-}\text{volume of gas; n}\text{-}\text{number of moles of gas}\\ \text{R}\text{-}\text{Gas constant; T - temperature in Kelvin}\end{array}$

Modify the ideal gas equation to calculate the number of moles of ${\text{H}}_{\text{2}}$

$\text{n}\text{=}\frac{\text{PV}}{\text{RT}}$

Substitute the values of P, V, R and T in the above equation

$\begin{array}{l}\text{mol}{\text{H}}_{\text{2}}\text{=}\frac{\left(\text{746}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)\left(\text{26}\text{.4}\text{L}\right)}{\left(\text{0}\text{.0821}\raisebox{1ex}{$\text{L}\cdot \text{atm}$}\!\left/ \!\raisebox{-1ex}{$\text{K}\cdot \text{mol}$}\right.\right)\left(\text{293}\text{K}\right)}\\ \text{=}\left(\frac{\text{0}\text{.9816}\text{×}\text{26}\text{.4}}{\text{0}\text{.0821}\text{×}\text{293}}\right)\text{mol}{\text{H}}_{\text{2}}\\ \text{=}\left(\frac{\text{25}\text{.9142}}{\text{24}\text{.0553}}\right)\text{mol}{\text{H}}_{\text{2}}\\ \text{=}\text{1}\text{.077}\text{mol}{\text{H}}_{\text{2}}\end{array}$

The number of moles of ${\text{H}}_{\text{2}}$ is calculated as 1.007.

By using the modified ideal gas equation and given data the number of moles hydrogen was calculated as 1.007 moles.

By using the balanced equation calculate the amount of ${\text{CaH}}_{\text{2}}$

${\text{CaH}}_{\text{2}}\left(\text{s}\right)\text{+}{\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\to \text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(\text{aq}\right)\text{+}{\text{2H}}_{\text{2}}\left(\text{g}\right)$

Two moles of ${\text{H}}_{\text{2}}$ produced from one mole of ${\text{CaH}}_{\text{2}}$.  So two moles of ${\text{CaH}}_{\text{2}}$ is equivalent to one mole of ${\text{H}}_{\text{2}}$

$\begin{array}{l}\text{mass}\text{of}{\text{CaH}}_{\text{2}}\text{=}\text{mol}{\text{H}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{CaH}}_{\text{2}}}{\text{2mol}{\text{H}}_{\text{2}}}\text{×}\frac{\text{molecular}\text{mass}\text{of}{\text{CaH}}_{\text{2}}}{\text{1}\text{mol}{\text{CaH}}_{\text{2}}}\\ \text{=}\text{1}\text{.077}\text{mol}{\text{H}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}{\text{CaH}}_{\text{2}}}{\text{2mol}{\text{H}}_{\text{2}}}\text{×}\frac{\text{42}\text{.0939}\text{g}}{\text{1}\text{mol}{\text{CaH}}_{\text{2}}}\\ \text{=}\left(\text{1}\text{.077}\text{×}\text{0}\text{.5}\text{×}\text{42}\text{.0939}\right)\text{g}\\ \text{=}\text{22}\text{.6676}\text{g}\end{array}$

The mass of ${\text{CaH}}_{\text{2}}$ needed to produce 26.4L of ${\text{H}}_{\text{2}}$ is 22.6676 g.

By using the balanced chemical equation and the number of moles of hydrogen the amount of calcium hydride needed was calculated as 22.6676g.

Conclusion

Balanced equation and the mass of calcium hydride needed to produce the required amount of hydrogen gas were determined.