One of the steps in the manufacture of nitric acid is the oxidation of ammonia, shown in this equation:

\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g) \]

If 55.0 kg of $\text{NH}_3$ reacts with 139 kg of $\text{O}_2$, what mass of $\text{NO}$ is formed?

\[ \_\_\_\_ \text{kg NO} \]

To find the mass of NO formed, we need to determine the limiting reactant (the reactant that is completely consumed) and then use stoichiometry to calculate the mass of NO produced. First, calculate the moles of NH₃ and O₂ given the masses provided:

Mass of NH₃ = 55.0 kg

Mass of O₂ = 139 kg

Molar mass of NH₃ = 17.03 g/mol

Molar mass of O₂ = 32.00 g/mol

Moles of NH₃ = (55.0 kg) / (17.03 g/mol) = 3234.2 mol

Moles of O₂ = (139 kg) / (32.00 g/mol) = 4343.8 mol

In the balanced equation, the ratio of NH₃ to NO is 4:4 or 1:1, meaning that for every mole of NH₃, 1 mole of NO is formed. Therefore, the number of moles of NO formed will be equal to the number of moles of NH₃.

Mass of NO formed = Moles of NO formed * Molar mass of NO

Mass of NO formed = 3234.2 mol * 30.01 g/mol = 97062.14 g

Therefore, the mass of NO formed is 97062.14 g or 97.06 kg.

Learn more about Stoichiometry, Limiting reactant, Mass calculation here:

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