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 H 2 O that can be produced by combining 65.8 g of each reactant? 4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Answer 1

Answer: 44.5 g of $H_2O$ will be produced from the given masses of both reactants.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$    

$\text{Moles of} NH_3=\frac{65.8}{17}=3.87moles$

$\text{Moles of} O_2=\frac{65.8}{32}=2.06moles$

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

According to stoichiometry :

5 moles of $O_2$ require 4 moles of $NH_3$

Thus 2.06 moles of $O_2$ will require=$\frac{4}{5}\times 2.06=1.65moles$  of $NH_3$

Thus $O_2$ is the limiting reagent as it limits the formation of product and $NH_3$ is the excess reagent.

As 5 moles of $O_2$ give = 6 moles of $H_2O$

Thus 2.06 moles of $O_2$ give =$\frac{6}{5}\times 2.06=2.47moles$  of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=2.47moles\times 18g/mol=44.5g$

Thus 44.5 g of $H_2O$ will be produced from the given masses of both reactants.