Chapter 8, Problem 8.38QP

Interpretation Introduction

Interpretation: The amount of ${\text{NH}}_{\text{3}}$ needed to produce $\text{1}\text{.00}\text{×}{\text{10}}^{\text{5}}\text{Kg}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be determined.

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:

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

To calculate: The amount of ${\text{NH}}_{\text{3}}$ needed to produce $\text{1}\text{.00}\text{×}{\text{10}}^{\text{5}}\text{Kg}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$

Answer to Problem 8.38QP

The amount of ${\text{NH}}_{\text{3}}$ needed to produce $\text{1}\text{.00}\text{×}{\text{10}}^{\text{5}}\text{Kg}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{2}\text{.58}\text{×}{\text{10}}^{\text{4}}\text{Kg}$

Explanation of Solution

The mass of products should be equal to mass of its starting materials this is governed by law of conservation of mass this equation is known as balanced chemical equation.

Ammonium sulphate production equation is given below:

${\text{2NH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}$

The above balanced equation shows 1 mole of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ produced from 2 moles of ${\text{NH}}_{\text{3}}$

$\frac{\text{1}\text{mole}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2}\text{moles}\text{of}{\text{NH}}_{\text{3}}}$

Molar mass of ${\left(N{H}_4\right)}_{2}S{O}_4$ calculation is given below:

$\begin{array}{l}\text{Molar}\text{mass}\text{of}\left({\text{NH}}_{\text{4}}\right){\text{SO}}_{\text{4}}\text{=}\left(\left(\text{2×mass}\text{of}\text{N}\right)\text{+}\left(\text{8×mass}\text{of}\text{H}\right)\text{+}\text{mass}\text{of}\text{S+}\left(\text{4×mass}\text{of}\text{O}\right)\right)\text{grams}\\ \text{=}\left(\left(\text{2×14}\text{.0067}\right)\text{+}\left(\text{8×1}\text{.00794}\right)\text{+}\text{32}\text{.065}\text{+}\left(\text{4×15}\text{.9994}\right)\right)\text{grams}\\ \text{=}\left(\text{28}\text{.0134}\text{+}\text{8}\text{.06352}\text{+}\text{32}\text{.065}\text{+}\text{63}\text{.9976}\right)\text{grams}\\ \text{=}\text{132}\text{.13952}\text{grams}\end{array}$Molar mass of $N{H}_3$ calculation is given below:

$\begin{array}{l}\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{3}}\text{=}\left(\text{mass}\text{of}\text{N}\text{+}\text{3×mass}\text{of}\text{H}\right)\text{grams}\\ \text{=}\left(\text{14}\text{.0067}\text{+}\text{3×1}\text{.00794}\right)\text{grams}\\ \text{=}\left(\text{14}\text{.0067+}\text{3}\text{.02382}\right)\text{grams}\\ \text{=}\text{17}\text{.03052}\text{grams}\end{array}$

Kilograms to moles conversion of ${\left(N{H}_4\right)}_{2}S{O}_4$

$\begin{array}{l}\frac{\text{Weight}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{Molecular}\text{weight}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}}\text{=}\text{moles}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\\ \frac{\text{100000000}}{\text{132}\text{.13952}}\text{=}\text{75675}\text{.8684}\text{moles}\end{array}$

This will require double the amount of ${\text{NH}}_{\text{3}}$

$\text{75675}\text{.8684}\text{×}\text{2}\text{=}\text{151351}\text{.7368}\text{moles}\text{of}{\text{NH}}_{\text{3}}$

The amount of $N{H}_3$ needed calculation is given below:

$\begin{array}{l}\text{amount}\text{of}{\text{NH}}_{\text{3}}\text{needed}\text{in}\text{grams}\text{=}\text{moles}\text{of}{\text{NH}}_{\text{3}}\text{×}\text{molecular}\text{weight}\text{of}{\text{NH}}_{\text{3}}\\ \text{=}\left(\text{151351}\text{.7368}\text{×}\text{17}\text{.03052}\right)\text{grams}\\ \text{=}\text{2577598}\text{.7806}\text{grams}\\ \text{=}\text{2}\text{.58}\text{×}{\text{10}}^{\text{7}}\text{grams}\\ \text{=}\text{2}\text{.58}\text{×}{\text{10}}^{\text{4}}\text{Kilograms}\end{array}$

The amount of ${\text{NH}}_{\text{3}}$ needed to produce $\text{1}\text{.00}\text{×}{\text{10}}^{\text{5}}\text{Kg}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{2}\text{.58}\text{×}{\text{10}}^{\text{4}}\text{Kg}$

Conclusion

The amount of ${\text{NH}}_{\text{3}}$ needed to produce $\text{1}\text{.00}\text{×}{\text{10}}^{\text{5}}\text{Kg}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ was determined.

 The amount of ${\text{NH}}_{\text{3}}$ needed to produce $\text{1}\text{.00}\text{×}{\text{10}}^{\text{5}}\text{Kg}\text{of}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{2}\text{.58}\text{×}{\text{10}}^{\text{4}}\text{Kg}$