Question

You conduct an experiment that requires the creation of an ammonia solution. You do this by reacting 50.0 L of nitrogen gas with excess hydrogen gas at STP to create ammonia gas using the Haber process given by the following reaction: N2 (g) + 3H2 (g) → 2NH3 (g). Assuming the reaction had 100% yield, you then bubble the ammonia into 5.0 L of water.
What is the molarity of the resulting ammonia solution?

Answer 1

Answer : The molarity of the resulting ammonia solution is, 0.89 M

Explanation :

The balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

First we have to calculate the moles of nitrogen gas.

As we know that at STP, 1 mole of gas occupies 22.4 L volume of gas.

As, 22.4 L volume of nitrogen gas present in 1 moles of nitrogen gas

So, 50.0 L volume of nitrogen gas present in $\frac{50.0}{22.4}=2.23$ moles of nitrogen gas

Thus, the moles of nitrogen gas is 2.23 moles.

Now we have to calculate the moles of ammonia gas.

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 2.23 moles of $N_2$ react to give $2.23\times 2=4.46$ moles of $NH_3$

Now we have to calculate the molarity of the resulting ammonia solution.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of ammonia}}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{4.46mole}{5.0L}$

$\text{Molarity}=0.89M$

Therefore, the molarity of the resulting ammonia solution is, 0.89 M