Chapter 8, Problem 8.96QP

Interpretation Introduction

Interpretation: for the given reaction, the masses of products formed from 30g of ${\text{NH}}_3$ and 175g of ${\text{ClF}}_{\text{3}}$ 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 reactant and a product of a reaction are depends upon the coefficients of reactant and product in a balanced chemical equation.
• Equation for Number of moles of a substance,

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

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

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

Answer to Problem 8.96QP

The mass of products formed in the given reaction are $\text{24}\text{.7g}{\text{N}}_{\text{2}}\text{,}\text{62}\text{.5g}{\text{Cl}}_{\text{2}}\text{and}\text{106g}\text{HF}$.

Explanation of Solution

In given reaction diagram,

2 molecules of ${\text{ClF}}_3$ reacts with two molecules of ${\text{NH}}_3$ to form 1 molecule of ${\text{N}}_2$, 1 molecule of ${\text{Cl}}_2$, two molecules of ${\text{ClF}}_3$ and six molecules of $\text{HF}$.

So, the chemical equation for this reaction is,

${\text{2ClF}}_3+2{\text{NH}}_3\to {\text{N}}_2+{\text{Cl}}_2+\text{6HF}$

Balanced chemical equation of a reaction is written according to law of conservation of mass.

Therefore,

The total number of each atoms in the reactant side should equal to the total number of each atoms in the product side.

So, in order to balance a chemical equation, the coefficients of compounds or atoms are needed to be changed in such a way that total number of each atoms in the reactant side and the total number of each atoms in the product side is to become equal.

Hence,

The above chemical equation is balanced;

${\text{2ClF}}_3+2{\text{NH}}_3\to {\text{N}}_2+{\text{Cl}}_2+\text{6HF}$

The mass of ${\text{ClF}}_3$ in the given reaction is given as $\text{175g}$.

The mass of ${\text{NH}}_3$ in the given reaction is given as $\text{30g}$.

Equation for 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 ${\text{ClF}}_3$ is,

$\text{Number of moles}\text{=}\frac{\text{175g}}{\text{92}\text{.45g}}\text{=}\text{1}\text{.893}\text{mol}$

The number of moles of ${\text{NH}}_3$ is,

$\text{Number of moles}\text{=}\frac{\text{30g}}{\text{17}\text{.03g}}\text{=}\text{1}\text{.762}\text{mol}$

The balanced equation of the reaction is found as,

${\text{2ClF}}_3+2{\text{NH}}_3\to {\text{N}}_2+{\text{Cl}}_2+\text{6HF}$

The mole ratio between reactants ${\text{ClF}}_3$ and ${\text{NH}}_3$ is 1:1.

The number of moles of ${\text{ClF}}_3$ is found as $\text{1}\text{.893}\text{mol}$.

The number of moles of ${\text{NH}}_3$ is found as $\text{1}\text{.762}\text{mol}$.

Therefore,

The limiting reactant of the reaction is ${\text{NH}}_3$, because it has less number of moles, so it consumes first in completely.

The balanced equation of the reaction is found as,

${\text{2ClF}}_3+2{\text{NH}}_3\to {\text{N}}_2+{\text{Cl}}_2+\text{6HF}$

The mole ratio between reactant ${\text{NH}}_3$ and product ${\text{N}}_2$ is $\text{1:}\frac{\text{1}}{\text{2}}$.

The number of moles of ${\text{NH}}_3$ is found as $\text{1}\text{.762}\text{mol}$.

Therefore,

The number of moles of ${\text{N}}_2$ formed is,

$\frac{\text{1}}{\text{2}}\text{×}\text{1}\text{.762}\text{mol}\text{=}\text{0}\text{.881}\text{mol}$

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

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

Therefore,

The mass of ${\text{N}}_2$ formed in the given reaction is,

$\text{0}\text{.881}\text{×}\text{28}\text{.02g}\text{=}\text{24}\text{.7g}$

The balanced equation of the reaction is found as,

${\text{2ClF}}_3+2{\text{NH}}_3\to {\text{N}}_2+{\text{Cl}}_2+\text{6HF}$

The mole ratio between reactant ${\text{NH}}_3$ and product ${\text{Cl}}_2$ is $\text{1:}\frac{\text{1}}{\text{2}}$.

The number of moles of ${\text{NH}}_3$ is found as $\text{1}\text{.762}\text{mol}$.

Therefore,

The number of moles of ${\text{Cl}}_2$ formed is,

$\frac{\text{1}}{\text{2}}\text{×}\text{1}\text{.762}\text{mol}\text{=}\text{0}\text{.881}\text{mol}$

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

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

Therefore,

The mass of ${\text{Cl}}_2$ formed in the given reaction is,

$\text{0}\text{.881}\text{×}\text{70}\text{.9g}\text{=}\text{62}\text{.5g}$

The balanced equation of the reaction is found as,

${\text{2ClF}}_3+2{\text{NH}}_3\to {\text{N}}_2+{\text{Cl}}_2+\text{6HF}$

The mole ratio between reactant ${\text{NH}}_3$ and product $\text{HF}$ is $\text{1:3}$.

The number of moles of ${\text{NH}}_3$ is found as $\text{1}\text{.762}\text{mol}$.

Therefore,

The number of moles of $\text{HF}$ formed is,

$\text{3}\text{×}\text{1}\text{.762}\text{mol}\text{=}\text{5}\text{.286}\text{mol}$

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

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

Therefore,

The mass of $\text{HF}$ formed in the given reaction is,

$\text{5}\text{.286}\text{×}2\text{0}\text{.01g}\text{=}10\text{6g}$

Conclusion

Conclusion

The masses of products formed from 30g of ${\text{NH}}_3$ and 175g of ${\text{ClF}}_{\text{3}}$ in the given reaction are determined according to the data’s given.