Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg}\) \(\mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction, and assume that the NO produced in the third step is not recycled.

Step by step solution

01

Analyzing the Ostwald process

The Ostwald process is represented by three chemical equations: 1. 4NH_3(g) + 5O_2(g) → 4NO(g) + 6H_2O(g) 2. 2NO(g) + O_2(g) → 2NO_2(g) 3. 3NO_2(g) + H_2O(l) → 2HNO_3(aq) + NO(g) Our goal is to find the mass of \(NH_3\) needed to produce \(1.0 \times 10^6 kg\) of \(HNO_3\). We'll start by finding the molar ratio between \(NH_3\) and \(HNO_3\).

02

Finding the total reaction by adding the equations

To find the net reaction of the Ostwald process, we can add the three equations together. Keep in mind that the NO produced in the third equation is not recycled, so we're not going to subtract it. After adding equations 1, 2, and 3, we get the net equation: \(4NH_3(g) + 5O_2(g) + H_2O(l) \rightarrow 2HNO_3(aq) + 6H_2O(g) + NO(g)\) Now we have the molar ratio between \(NH_3\) and \(HNO_3\).

03

Calculate moles of HNO3 produced

To find the mass of \(NH_3\) required, we first need to calculate the moles of \(HNO_3\) produced: Let's define m_HNO3 = \(1.0 \times 10^6 kg\) /(\rho_HNO3) to transform the mass into moles /[M_HNO3] where \(\rho_{HNO3}\) is the density of \(HNO_3\) and \[M_{HNO3} = N_a \times \left(1 \times m_H + 1 \times m_N + 3 \times m_O\right)\] Insert the atomic masses here: \[M_{HNO3} = N_a \times \left(1 \times 1 + 1 \times 14 + 3 \times 16\right)\] Moles of HNO3 produced = \(n_{HNO3} = \frac{m_{HNO3}}{M_{HNO3}}\)

04

Calculate moles of NH3 needed and find the mass

From the net equation, the molar ratio between \(NH_3\) and \(HNO_3\) is 4:2, so for each 2 moles of \(HNO_3\) produced, we need 4 moles of \(NH_3\). To find the moles of NH3 required, calculate: \(n_{NH3} = 2 \times n_{HNO3}\) Now we can find the mass of \(NH_3\) required using the moles and the molar mass of \(NH_3\): \[M_{NH3} = N_a \times \left(1 \times m_N + 3 \times m_H\right)\] Insert the atomic masses here : \[M_{NH3} = N_a \times \left(1 \times 14 + 3 \times 1\right)\] Mass of NH3 needed = \(m_{NH3} = n_{NH3} \times M_{NH3}\) Now you can find the mass of \(NH_3\) required to produce \(1.0 \times 10^6 kg\) of \(HNO_3\).

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