Chapter 3, Problem 3.97QP

Interpretation Introduction

Interpretation: for the given Ostwald process, the mass of reactant ${\text{NH}}_3$ required to produce 1 ton of ${\text{HNO}}_{\text{3}}$ to be determined.

Concept introduction:

• Balanced chemical equation of a reaction is written according to law of conservation of mass.
• Mole ratio between the reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.
• Number of moles of a substance,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

• Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Answer to Problem 3.97QP

The mass of ${\text{NH}}_3$ required to produce 1 ton of ${\text{HNO}}_{\text{3}}$ is $\text{9}\text{.58}\text{×}{\text{10}}^{\text{5}}\text{g}$.

Explanation of Solution

The mass of produced ${\text{HNO}}_{\text{3}}$ is given as,

$\text{1}\text{ton}\text{=}\frac{\text{2000}\text{lb}}{\text{1}\text{ton}}\text{=}\text{2000}\text{lb}\text{×}\frac{\text{453}\text{.6g}}{\text{1}\text{lb}}\text{=}\text{907200g}$

Equation for finding Number of moles of a substance,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Therefore,

The number of moles of produced ${\text{HNO}}_{\text{3}}$ is,

$\frac{\text{907200g}}{\text{63}\text{.018g}}\text{=}\text{1}\text{.44}{\text{×10}}^{\text{4}}\text{mol}$

Hence,

The number of moles of produced ${\text{HNO}}_{\text{3}}$ is $\text{1}\text{.44}{\text{×10}}^{\text{4}}\text{mol}$.

The balanced equation of the reaction of formation of ${\text{HNO}}_{\text{3}}$ is given as,

${\text{2NO}}_2\text{+}{\text{H}}_{\text{2}}\text{O}\to {\text{HNO}}_3+{\text{HNO}}_2$

The mole ratio between reactant ${\text{NO}}_2$ and product ${\text{HNO}}_{\text{3}}$ is 2:1.

That is,

Two moles of${\text{NO}}_2$ molecules reacts to form the 1 mole of ${\text{HNO}}_{\text{3}}$ molecules.

The number of moles of produced ${\text{HNO}}_{\text{3}}$ is found as $\text{1}\text{.44}{\text{×10}}^{\text{4}}\text{mol}$.

Therefore,

The number of moles of ${\text{NO}}_2$ required to produce $\text{1}\text{.44}{\text{×10}}^{\text{4}}\text{mol}$ of ${\text{HNO}}_{\text{3}}$ is,

$\text{2}\text{×}\text{1}\text{.44}{\text{×10}}^{\text{4}}\text{mol}\text{=}\text{2}{\text{.88×10}}^{\text{4}}\text{mol}$

But the percent yield of this reaction is 80%, so the number of moles of ${\text{NO}}_2$ required should be 100/80 of this amount.

Hence,

The number of moles of ${\text{NO}}_2$ required to produce $\text{1}\text{.44}{\text{×10}}^{\text{4}}\text{mol}$ of ${\text{HNO}}_{\text{3}}$ is,

$\text{2}{\text{.88×10}}^{\text{4}}\text{mol}\text{×}\frac{\text{100\%}}{\text{80\%}}\text{=}\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}$

The balanced equation of the reaction of formation of ${\text{NO}}_2$ is given as,

$\text{2NO}\text{+}{\text{O}}_2\to 2{\text{NO}}_2$

The mole ratio between reactant $\text{NO}$ and product ${\text{NO}}_2$ is 1:1.

That is,

1 mole of$\text{NO}$ molecules reacts to form the 1 mole of ${\text{NO}}_2$ molecules.

The number of moles of produced ${\text{NO}}_2$ is found as $\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}$.

Therefore,

The number of moles of $\text{NO}$ required to produce $\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}$ of ${\text{NO}}_2$ is,

$\text{1}\text{×}\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}\text{=}\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}$

But the percent yield of this reaction is 80%, so the number of moles of $\text{NO}$ required to produce $\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}$ of ${\text{NO}}_2$ should be 100/80 of this amount.

Hence,

The number of moles of $\text{NO}$ required to produce $\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}$ of ${\text{NO}}_2$ is,

$\text{3}\text{.6}{\text{×10}}^{\text{4}}\text{mol}\text{×}\frac{\text{100\%}}{\text{80\%}}\text{=}\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}$

The balanced equation of the reaction of formation of $\text{NO}$ is given as,

${\text{4NH}}_3\text{+}5{\text{O}}_2\to 4\text{NO}\text{+}6{\text{H}}_2\text{O}$

The mole ratio between reactant ${\text{NH}}_3$ and product $\text{NO}$ is 1:1.

That is,

1 mole of${\text{NH}}_3$ molecules reacts to form the 1 mole of $\text{NO}$ molecules.

The number of moles of produced $\text{NO}$ is found as $\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}$.

Therefore,

The number of moles of ${\text{NH}}_3$ required to produce $\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}$ of $\text{NO}$ is,

$\text{1}\text{×}\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}\text{=}\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}$

But the percent yield of this reaction is 80%, so the number of moles of ${\text{NH}}_3$ required to produce $\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}$ of $\text{NO}$ should be 100/80 of this amount.

Hence,

The number of moles of ${\text{NH}}_3$ required to produce $\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}$ of $\text{NO}$ is,

$\text{4}\text{.5}{\text{×10}}^{\text{4}}\text{mol}\text{×}\frac{\text{100\%}}{\text{80\%}}\text{=}\text{5}\text{.625}{\text{×10}}^{\text{4}}\text{mol}$

The number of moles of ${\text{NH}}_3$ required to produce 1 ton of ${\text{HNO}}_{\text{3}}$ is found as $\text{5}\text{.625}{\text{×10}}^{\text{4}}\text{mol}$.

Equation for finding Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

That is,

$\text{5}\text{.625}{\text{×10}}^{\text{4}}\text{mol}\times \text{17}\text{.034}\text{g}\text{=}\text{9}\text{.58}\text{×}{\text{10}}^{\text{5}}\text{g}$

Conclusion

The mass of reactant ${\text{NH}}_3$ required to produce 1 ton of ${\text{HNO}}_{\text{3}}$ is determined according to the data’s given.