Question

The balanced chemical equation for the formation of ammonia gas by the reaction between nitrogen gas and hydrogen gas is given
$N_2+3H_2\to 2N{H}_3$
How many moles of ammonia are formed when $6$ moles of $N_2$ react with $6$ moles of $H_2$?

Answer 1

Hint: The limiting reagent is the one which is consumed in the reaction process when the reaction is completed. It limits the amount of product which can be formed. So we need to find the limiting reagent in this reaction and consequently the moles of ammonia needed to react with $6$ moles of $H_2$

Complete step by step solution: We have been given the balanced chemical equation for the formation of ammonia which is stated as follows
$N_2+3H_2\to 2N{H}_3$
From the above reaction we infer that $1$ mole of $N_2$ reacts with $3$ moles of $H_2$
Therefore $6$ moles of will react with 18 moles of $H_2$
Now the given amount of $H_2$ = 6 moles
Therefore $H_2$ is the limiting reagent as it is comp$N_2$letely consumed in the reaction.
Now again from the given reaction we infer that
Number of moles of ammonia formed when $3$ moles of $H_2$ reacts with excess of $N_2$=$2$
$\therefore$Number of moles of ammonia formed when $6$ moles of $H_2$ reacts with excess of $N_2$=${\displaystyle \frac{2}{3}}\times 6=4$ moles
Therefore $4$ moles of ammonia are formed when $6$ moles of $N_2$ reacts with $6$ moles of $H_2$

Therefore the number of moles of ammonia formed is four.

Note: Limiting reagent can be found by the following procedures
-Determining the balanced chemical equation
-Convert all the given information into moles
-Calculate the mole ratio from the given data.
-Use the amount of limiting reagent to find the amount of product formed.
These are the procedures to find the limiting reagent in a chemical reaction.