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)](https://tex.z-dn.net/?f=N_2%28g%29%2B3H_2%28g%29%5Crightarrow%202NH_3%28g%29)
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
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
react to give 2 mole of ![NH_3](https://tex.z-dn.net/?f=NH_3)
So, 2.23 moles of
react to give
moles of ![NH_3](https://tex.z-dn.net/?f=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)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20ammonia%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Now put all the given values in this formula, we get:
![\text{Molarity}=\frac{4.46mole}{5.0L}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%7D%3D%5Cfrac%7B4.46mole%7D%7B5.0L%7D)
![\text{Molarity}=0.89M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%7D%3D0.89M)
Therefore, the molarity of the resulting ammonia solution is, 0.89 M