Question

“There is a mismatch between man’s ability to produce and social incapacity to utilize”.
Explain the meaning of this statement

Answer 1

Question#46528 – Chemistry – Inorganic Chemistry

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 $\mathrm{H}_2\mathrm{O}$ that can be produced by combining 86.4 g of each reactant?

$$4 \mathrm{NH}_{3(\mathrm{g})} + 5 \mathrm{O}_{2(\mathrm{g})} \rightarrow 4 \mathrm{NO}_{(\mathrm{g})} + 6 \mathrm{H}_{2}\mathrm{O}_{(\mathrm{g})}$$

Answer:

Firstly, the amount of moles of each initial compound can be calculated:

$$\begin{array}{l} n\left(NH_3\right) = \frac{m\left(NH_3\right)}{M\left(NH_3\right)} = \frac{86.4 \mathrm{g}}{17 \mathrm{g/mol}} = 5.1 \mathrm{mol} \\ n(O_2) = \frac{m(O_2)}{M(O_2)} = \frac{86.4 \mathrm{g}}{32 \mathrm{g/mol}} = 2.7 \mathrm{mol} \\ \end{array}$$

From the equation above it is obvious, that the theoretical molar ratio between $\mathrm{NH}_3$ and $\mathrm{O}_2$ is 4:5. In our case there is $n(\mathrm{NH}_3):n(\mathrm{O}_2) = 5.1:2.7$.

Therefore, the limiting reactant is oxygen gas and ammonia is in excess. And we will continue calculations using amount of moles of $\mathrm{O}_2$.

If 5 moles of $\mathrm{O}_2$ produce 6 moles of water $\mathrm{H}_2\mathrm{O}$, than 2.7 moles of $\mathrm{O}_2$ produce x moles of $\mathrm{H}_2\mathrm{O}$. Using the following proportion, one can calculate x:

$$\frac{5}{6} = \frac{2.7}{x} \Rightarrow x = \frac{2.7 \times 6}{5} = 3.24 \mathrm{mol}$$

The mass of water can be calculated considering that mass of one mole of water is $18 \mathrm{~g/mol}$:

$$m(H_2O) = n(H_2O) \times M(H_2O) = 3.24 \mathrm{mol} \times 18 \mathrm{g/mol} = 58.32 \mathrm{g}$$