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 step is recycled?
kg nh₃
an error has been detected in your answer. check for typos, miscalculations etc. before submitting your answer.

Explanation:

Step1: Find the molar ratio of \(NH_3\) to \(HNO_3\)

From the three - step Ostwald process reactions:

• In the first reaction \(4NH_3(g)+5O_2(g)\to4NO(g) + 6H_2O(g)\)
• In the second reaction \(2NO(g)+O_2(g)\to2NO_2(g)\)
• In the third reaction \(3NO_2(g)+H_2O(l)\to2HNO_3(aq)+NO(g)\)

Combining these reactions (by multiplying the second reaction by \(3\) and the third reaction by \(2\) and then adding to the first reaction and canceling out the intermediate \(NO\) and \(NO_2\) species), we get a net ratio of \(NH_3\) to \(HNO_3\) as \(1:1\) (after proper stoichiometric balancing and canceling of intermediates).

Step2: Calculate the moles of \(HNO_3\)

The mass of \(HNO_3\) is \(m = 2.0\times10^{6}\space kg=2.0\times 10^{9}\space g\). The molar mass of \(HNO_3\) is \(M_{HNO_3}=1 + 14+3\times16=63\space g/mol\). The number of moles of \(HNO_3\), \(n_{HNO_3}=\frac{m}{M}=\frac{2.0\times 10^{9}\space g}{63\space g/mol}\approx3.17\times 10^{7}\space mol\)

Step3: Calculate the mass of \(NH_3\)

Since the mole ratio of \(NH_3\) to \(HNO_3\) is \(1:1\), the number of moles of \(NH_3\), \(n_{NH_3}=n_{HNO_3}\approx3.17\times 10^{7}\space mol\). The molar mass of \(NH_3\) is \(M_{NH_3}=14 + 3\times1=17\space g/mol\). The mass of \(NH_3\), \(m_{NH_3}=n_{NH_3}\times M_{NH_3}=3.17\times 10^{7}\space mol\times17\space g/mol = 5.39\times 10^{8}\space g = 5.4\times 10^{5}\space kg\)

Answer:

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