Question

nitric acid is produced commercially by the ostwald process. the three steps of the ostwald process are shown in the following equations:
4nh₃(g) + 5o₂(g) → 4no(g) + 6h₂o(g)
2no(g) + o₂(g) → 2no₂(g)
3no₂(g) + h₂o(l) → 2hno₃(aq) + no(g)
what mass of nh₃ must be used to produce 2.0×10⁶ kg of hno₃ by the ostwald process, assuming a 100% yield in each reaction and assuming the no produced in the third stage is not recycled?
kg nh₃

Explanation:

Step1: Combine the reactions

From the three reactions:
$$4NH_{3}(g)+5O_{2}(g)\to4NO(g)+6H_{2}O(g)$$
$$2NO(g)+O_{2}(g)\to2NO_{2}(g)$$
$$3NO_{2}(g)+H_{2}O(l)\to2HNO_{3}(aq)+NO(g)$$
Multiply the second reaction by \(3\) and the third reaction by \(2\), then add them together with the first reaction. After canceling out the intermediate \(NO\) and \(NO_{2}\), we get the overall reaction:
$$4NH_{3}(g)+8O_{2}(g)\to4HNO_{3}(aq)+4H_{2}O(g)$$
Simplify to \(NH_{3}(g)+2O_{2}(g)\to HNO_{3}(aq)+H_{2}O(g)\)

Step2: Calculate the molar mass

The molar mass of \(NH_{3}\) is \(M_{NH_{3}}=(14 + 3\times1)\space g/mol=17\space g/mol\)
The molar mass of \(HNO_{3}\) is \(M_{HNO_{3}}=(1 + 14+3\times16)\space g/mol = 63\space g/mol\)

Step3: Set up the proportion

Let \(m\) be the mass of \(NH_{3}\). According to the stoichiometry of the reaction \(NH_{3}(g)+2O_{2}(g)\to HNO_{3}(aq)+H_{2}O(g)\), the ratio of \(n(NH_{3})\) to \(n(HNO_{3})\) is \(1:1\). So \(\frac{m}{M_{NH_{3}}}=\frac{m_{HNO_{3}}}{M_{HNO_{3}}}\)
Given \(m_{HNO_{3}} = 2.0\times10^{6}\space kg=2.0\times10^{9}\space g\)
Substitute into the formula: \(m=\frac{M_{NH_{3}}}{M_{HNO_{3}}}\times m_{HNO_{3}}\)

Step4: Calculate the mass of \(NH_{3}\)

\(m=\frac{17\space g/mol}{63\space g/mol}\times2.0\times 10^{9}\space g\approx5.4\times10^{8}\space g = 5.4\times10^{5}\space kg\)

Answer:

\(5.4\times 10^{5}\space kg\)