Question

$56.0 \,L$ of nitrogen gas is mixed with excess of hydrogen gas and it is found that $20\, L$ of ammonia gas is produced The volume of unused nitrogen gas is found to be _______ $L$

Answer 1

Correct Answer: 46

To solve the problem, let's first understand the chemical reaction involved. The balanced equation for the formation of ammonia from nitrogen and hydrogen gases is:

\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]

This equation tells us that 1 volume of nitrogen reacts with 3 volumes of hydrogen to produce 2 volumes of ammonia.

Given: The initial volume of nitrogen gas is 56.0 L, and 20 L of ammonia is produced.

Using the stoichiometry of the reaction, for every 2 volumes of ammonia produced, 1 volume of nitrogen is consumed. Therefore, the volume of nitrogen used is calculated as follows:

Volume of ammonia produced = 20 L

According to the balanced equation, 2 L of ammonia is produced from 1 L of nitrogen. So, the volume of nitrogen gas used is:

Volume of nitrogen used = \(\frac{20\,L}{2} = 10\,L\)

Now, calculate the unused volume of nitrogen gas by subtracting the volume of nitrogen used from the initial volume:

Volume of unused nitrogen = Initial volume - Volume of nitrogen used = 56.0 L - 10 L = 46.0 L

The volume of unused nitrogen gas is 46.0 L, which falls within the given range of 46 to 46, confirming the calculation is correct.