Question

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2OH2O that can be produced by combining 70.170.1 g of each reactant?

Answer 1

Answer:

$\boxed{\text{47.4 g}}$

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    17.03   32.00                 18.02  

           4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g:     70.1      70.1

Step 1. Calculate the moles of each reactant

$\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H _{2} O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}$

Step 2. Identify the limiting reactant  

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

$\text{Moles of H _{2} O} = \text{4.116 mol NH _{3} } \times \dfrac{\text{6 mol H _{2} O}}{\text{4 mol NH _{3} }} = \text{6.174 mol H _{2} O}$

From O₂:  

The molar ratio of H₂O:O₂ is 6:5.  

$\text{Moles of H _{2} O} = \text{2.191 mol O _{2} } \times \dfrac{\text{6 mol H _{2} O}}{\text{5 mol O _{2} }} = \text{2.629 mol H _{2} O}$

O₂ is the limiting reactant because it gives the smaller amount of H₂O.  

Step 3. Calculate the theoretical yield.

$\text{Theor. yield } = \text{2.629 mol H _{2} O}\times \dfrac{\text{18.02 g H _{2} O}}{\text{1 mol H _{2} O}} = \textbf{47.4 g H _{2} O}\\\\\text{The maximum yield of H _{2} O is }\boxed{\textbf{47.4 g}}$