Chapter 8.3, Problem 8.6WE

(a)

Interpretation Introduction

Interpretation: for the given reaction, the mass of ammonium nitrate required to produce $\text{10}\text{g}$ of nitrous oxide and the mass of water formed are needed to be determined.

Concept introduction:

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

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

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

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

To find: the number of moles of nitrous oxide formed in the given reaction.

Explanation of Solution

The mass of nitrous oxide is given as $\text{10}\text{g}$.

Equation for Number of moles of a substance, from its given mass is,

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

Therefore,

The number of moles of nitrous oxide formed in the given reaction is,

$\text{Number of moles}\text{=}\frac{\text{10}\text{g}}{\text{44}\text{.02g}}\text{=}\text{0}\text{.227}\text{mol}$

The balanced equation of the reaction is given as,

${\text{NH}}_4{\text{NO}}_{\text{3}}\to {\text{N}}_2\text{O}+2{\text{H}}_2\text{O}$

The mole ratio between reactant ${\text{NH}}_4{\text{NO}}_{\text{3}}$ and product ${\text{N}}_2\text{O}$ is 1:1.

The number of moles of ${\text{N}}_2\text{O}$ in the given reaction is found as $\text{0}\text{.227}\text{mol}$.

Therefore,

The number of moles of hydrogen required to react with $\text{0}\text{.0880}\text{mol}$ of nitrogen in the given reaction is $\text{1}\text{×}\text{0}\text{.227}\text{mol}\text{=}\text{0}\text{.227}\text{mol}$

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

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

Then,

The mass of ammonium nitrate (${\text{NH}}_4{\text{NO}}_{\text{3}}$) is,

$\text{0}\text{.227}\times \text{80}\text{.05}\text{g}\text{=}\text{18}\text{.2}\text{g}$

(b)

Interpretation Introduction

Interpretation: for the given reaction, the mass of ammonium nitrate required to produce $\text{10}\text{g}$ of nitrous oxide and the mass of water formed are needed to be determined.

Concept introduction:

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

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

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

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

To find: the mass of water formed in the given reaction.

Explanation of Solution

The balanced equation of the reaction is given as,

${\text{NH}}_4{\text{NO}}_{\text{3}}\to {\text{N}}_2\text{O}+2{\text{H}}_2\text{O}$

The mole ratio between products ${\text{N}}_2\text{O}$ and ${\text{H}}_2\text{O}$ is 1:2.

The number of moles of ${\text{N}}_2\text{O}$ in the given reaction is found as $\text{0}\text{.227}\text{mol}$.

Therefore,

The number of moles of ${\text{H}}_2\text{O}$ formed in the given reaction is $\text{2}\text{×}\text{0}\text{.227}\text{mol}\text{=}\text{0}\text{.454}\text{mol}$

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

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

Then,

The mass of ${\text{H}}_2\text{O}$ formed is,

$\text{0}\text{.227}\times \text{18}\text{.02}\text{g}\text{=}\text{8}\text{.18}\text{g}$

Conclusion

Conclusion

The mass of ammonium nitrate required to produce $\text{10}\text{g}$ of nitrous oxide and the mass of water formed are determined according to the given data’s.