Question

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multistep Ostwald process, which begins with the oxidation of ammonia:

Step 1.$\mathbf{4}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{5}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{4}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

Step 2. $\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Step 3. $\mathbf{3}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

(a) What are the oxidizing and reducing agents in each step?

(b) Assuming 100% yield in each step, what mass (in kg) of ammonia must be used to produce 3.0 X10⁴ kg of HNO₃?

Short answer

Answer of subpart (a):

Answer: You need to identify the oxidizing and reducing agents in each steps.

Answer of subpart (b):

Answer: You need to calculate the mass (in kg) of ammonia used to produce 3.010⁴ kg of HNO₃.

Step-by-step solution

Oxidation nos. for step-1 process 

Oxidation no of N in NH₃ is -3.

Oxidation no N in NO is +2.

Oxidation no ofelement oxygen (O₂) is 0.

Oxidation no of oxygen in H₂O is -2.

Conclusion

Oxidation no of N in NH₃ is -3 increases to +2 in NO. Hence, NH₃ undergoes oxidation and acts as a reducing agent.

Oxidation no of O in O₂ is 0 decreases to -2 in H₂O. Hence, O₂ undergoes reduction and acts as an oxidizing agent.

Oxidation nos. for step-2 process

Oxidation no of N in NO is +2.

Oxidation no N in NO₂ is +4.

Oxidation no ofelement oxygen (O₂) is 0.

Oxidation no of oxygen in NO₂ is -4.

Conclusion

Oxidation no of N in NO is +2 increases to +2 in NO₂. Hence, NO undergoes oxidation and acts as a reducing agent.

Oxidation no of O in O₂ is 0 decreases to -4 in NO₂. Hence, O₂ undergoes reduction and acts as an oxidizing agent.

Oxidation nos. for step-3 process

Oxidation no of N in NO₂ is +4.

Oxidation no N in NO is +2.

Oxidation no of N in HNO₃ is +5.

Conclusion

Oxidation no of N in NO₂ is +4 increases to +5 in HNO₃. Hence, NO₂ undergoes oxidation and acts as a reducing agent.

Oxidation no of N in NO₂ is +4 decreases to +2 in NO. Hence, NO₂ undergoes reduction and acts as an oxidizing agent.

Answer of subpart (b):

Answer: You need to calculate the mass (in kg) of ammonia used to produce 3.010⁴ kg of HNO₃.

Calculation of moles of HNO3

Mass of HNO₃ = 3.010⁴ kg = 3.010⁷ g

Molecular mass of HNO₃ = 63.008g/mol

Again you know,

$\mathbf{moles}\mathbf{=}\frac{\mathbf{mass}\mathbf{\left(}\mathbf{given}\mathbf{\right)}}{\mathbf{mass}\mathbf{\left(}\mathbf{molar}\mathbf{\right)}}$

Moles of HNO₃

_{$\begin{array}{l}\mathbf{=}\frac{\mathbf{3}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}}{\mathbf{63}\mathbf{.008}}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\\ \mathbf{=}\mathbf{0}\mathbf{.0476}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\end{array}$}

Hence, moles of HNO₃ produced are 0.047 $\mathbf{\times }$610⁷mol.

Calculation of moles of NO2

According to the balanced equation

$\mathbf{3}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{HNO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

3 moles of NO₂ forms 2 moles HNO₃

Now, moles of NO₂

^{role="math" localid="1656767569013" $\begin{array}{l}\mathbf{=}\left[(0.0476\times {10}^7)\times \frac{3}{2}\right]\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\\ \mathbf{=}\mathbf{0}\mathbf{.0714}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\end{array}$}

Hence, moles of HNO₃ are 0.0714 x 10⁷mol.

Calculation of moles of NO

According to the balanced equation

$\mathbf{2}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

2 moles of NO forms 2 moles NO₂

Now, moles of NO

$\begin{array}{l}\mathbf{=}\left[(0.0714\times {10}^7)\times \frac{2}{2}\right]\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\\ \mathbf{=}\mathbf{0}\mathbf{.0714}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\end{array}$

Hence, moles of NO are 0.0714 x 10⁷mol.

Calculation of moles of NH3

According to the balanced equation

$\mathbf{4}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{5}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{4}\mathbf{NO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

4 moles of NH₃ forms 4 moles NO

Now, moles of NH₃

^{$\begin{array}{l}\mathbf{=}\mathbf{[}\mathbf{(}\mathbf{0}\mathbf{.0714}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{)}\mathbf{\times }\frac{\mathbf{4}}{\mathbf{4}}\mathbf{]}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\\ \mathbf{=}\mathbf{0}\mathbf{.0714}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\end{array}$}

Hence, moles of NH₃ are 0.0714 x 10⁷mol.

Calculation of mass of NH3

Moles of NH₃ = 0.071410⁷mol.

Molecular mass of NH₃ = 17.024g/mol

Again you know,

$\mathbf{mass}\mathbf{=}\mathbf{moles}\mathbf{\times }\mathbf{mass}\mathbf{\left(}\mathbf{molar}\mathbf{\right)}$

Mass of NH₃

_{$\begin{array}{l}\mathbf{=}\mathbf{0}\mathbf{.0714}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\times }\mathbf{17}\mathbf{.024}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{=}\mathbf{1}\mathbf{.2155}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{=}\mathbf{1}\mathbf{.2155}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\left(}\mathbf{kg}\mathbf{\right)}\end{array}$}

Hence, mass of NH₃ is 1.2155 x 10⁴kg.

Conclusion

Hence, the mass of ammonia used to produce 3.010⁴ kg of HNO₃ is 1.2155 x 10⁴kg.