Question

An intermediate step in the industrial production of nitric acid involves the reaction of ammonia with oxygen gas to form nitrogen monoxide and water. How many grams of nitrogen monoxide can form by the reaction of 485 g of ammonia with 792 g of oxygen?

Short answer

Mass formed of NO during the reaction is 594 g.

Step-by-step solution

Writing balanced equation

First of all, let us check the balanced equation to find the amount of NO:

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

From the equation we can say that 4 molesof ammonia and 5 moles of oxygen gasreact to givefourmoles of NO gas.

Identifying Limiting Reagent

First of all, moles of reactants to be calculated to know about the limiting reagent. The moles of${\mathbf{NH}}_{\mathbf{3}}$ can be calculated as:

$\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}{\mathbf{Molar}\mathbf{}\mathbf{Mass}}\mathbf{=}\frac{\mathbf{485}}{\mathbf{17}}\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{5}$

The moles of${\mathbf{O}}_{\mathbf{2}}$can be calculated as:

$\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}{\mathit{O}}_2\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}{\mathit{O}}_{\mathbf{2}}}{\mathbf{Molar}\mathbf{}\mathbf{Mass}}\mathbf{=}\frac{\mathbf{792}}{\mathbf{32}}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{75}$

From the balanced equation we can say that ${\mathbf{O}}_{\mathbf{2}}$is the limiting reagent hence controls the formation of products and,

5 moles of ${\mathbf{O}}_{\mathbf{2}}$ gives moles of NO = 4 moles

So, 24.75 moles of will give = $\frac{\mathbf{4}}{\mathbf{5}}\mathbf{\times }\mathbf{24}\mathbf{.}\mathbf{75}\mathbf{=}\mathbf{19}\mathbf{.}\mathbf{8}\mathbf{Moles}$

Calculate mass of NO formed

The mass of NO formed is calculated as:

$\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{NO}\mathbf{=}\mathbf{Mole}\mathbf{\times }\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{=}\mathbf{19}\mathbf{.}\mathbf{8}\mathbf{\times }\mathbf{30}\mathbf{=}\mathbf{594}\mathbf{g}$

So, mass formed of NO during the reaction is 594 g.