Question

Nitric acid is produced commercially by the Ostwald process, represented by the following equations:

$$\begin{gathered} {\mathbf{\text{4NH}}}_{\mathbf{\text{3}}}{\mathbf{\text{(g) + 5O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{4NO(g) + 6H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(g)}} \\ {\mathbf{\text{2NO(g) +O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{2NO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}} \\ {\mathbf{\text{3NO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) +H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}\mathbf{\to }{\mathbf{\text{2HNO}}}_{\mathbf{\text{3}}}\mathbf{\text{(aq) + NO(g)}} \end{gathered}$$

What mass of${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$must be used to produce${\mathbf{\text{1.0×10}}}^{\mathbf{\text{6}}}$kg${\mathbf{\text{HNO}}}_{\mathbf{\text{3}}}$by the Ostwald process? Assume$\mathbf{\text{100\%}}$yield in each reaction, and assume that theproduced in the third step is not recycled.

Short answer

The mass of ${\text{NH}}_{\text{3}}$must be used to produce${\text{1.0×10}}^6$ kg ${\text{HNO}}_{\text{3}}$of by the Ostwald process is $N{H}_3={\text{40.1×10}}^{\text{7}}\text{g}$.

Step-by-step solution

Concept Introduction

The Ostwald method is a chemical method for producing nitric acid.

Balanced equations for Nitric Acid

Balanced reaction equations for the Ostwald process are shown below.

$$\begin{gathered} {\text{4NH}}_{\text{3}}{\text{(g) + 5O}}_{\text{2}}\text{(g)}\to {\text{4NO(g) + 6H}}_{\text{2}}\text{O(g)} \\ {\text{2NO(g) +O}}_{\text{2}}\text{(g)}\to {\text{2NO}}_{\text{2}}\text{(g)} \\ {\text{3NO}}_{\text{2}}{\text{(g) +H}}_{\text{2}}\text{O(l)}\to {\text{2HNO}}_{\text{3}}\text{(aq) + NO(g)} \end{gathered}$$

The yield of the product is determined with the help of limiting reagent.

Use the mole concept of a balanced chemical equation to determine the yield of the product in the chemical reaction.

Now, calculate the mass of${\text{NH}}_{\text{3}}$ needed to produce${\text{1.0×10}}^6$kg of nitric acid.

Mass of Nitric Acid

Moles of${\text{HNO}}_{\text{3}}$is calculated from the equation shown below –

$$\begin{gathered} \text{Moles =}\frac{{\text{mass of HNO}}_{\text{3}}}{\text{Molar mass}} \\ \text{=}\frac{{\text{1.0×10}}^{\text{6}}{\text{×10}}^{\text{3}}\text{g}}{\text{63.0g/mol}} \\ {\text{Moles HNO}}_{\text{3}}{\text{= 1.58×10}}^{\text{7}}\text{moles} \end{gathered}$$

The coefficients in a balanced chemical equation represent the reacting rations of the substances in the reaction.When the moles of one substance in a reaction are known, the coefficients of the balanced equation can be used todetermine the moles of all the substances in a reaction.

Therefore, the moles of${\text{NH}}_{\text{3}}$ can be calculated as follows –

$$\begin{gathered} {\text{Moles of NH}}_{\text{3}}{\text{= 1.58×10}}^{\text{7}}\text{moles×}\frac{{\text{3 moles NO}}_{\text{2}}}{\text{2 moles HNO}}\text{×}\frac{\text{2 moles NO}}{{\text{2 moles NO}}_{\text{2}}}\text{×}\frac{{\text{4 moles NH}}_{\text{3}}}{\text{4 moles NO}} \\ {\text{Moles NH}}_{\text{3}}{\text{= 2.37×10}}^{\text{7}}\text{moles} \end{gathered}$$

Finally, calculate mass of ${\text{NH}}_{\text{3}}$–

role="math" localid="1663823604887" $$\begin{gathered} {\text{Mass of NH}}_{\text{3}}{\text{= Moles of NH}}_{\text{3}}{\text{×Molar mass of NH}}_{\text{3}} \\ {\text{Mass of NH}}_{\text{3}}{\text{= 2.37×10}}^{\text{7}}\text{×17.0g/mol} \\ {\text{Mass of NH}}_{\text{3}}{\text{= 40.1×10}}^{\text{7}}\text{g} \end{gathered}$$

Therefore, the mass of ${\mathrm{NH}}_{\text{3}}{\text{= 40.1×10}}^{\text{7}}\text{g}$.