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

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Solution

Given reaction for the formation of ammonia-
$N_{2} + 3 H_{2} ⟶ 2 N H_{3}$
From the above reacton,
$1$ mole of $N_{2}$ reacts with $3$ moles of $H_{2}$ .
Therefore,
$6$ moles of $N_{2}$ will react with $18$ moles of $H_{2}$ .
Given amount of $H_{2} = 6 moles$
Therefore, $H_{2}$ is the limiting reagent.
Therefore,
Again from the above reaction.
No. of moles of ammonia formed when $3$ moles of $H_{2}$ react with excess of $N_{2} = 2$
$∴$ No. of moles of ammonia formed when $6$ moles of $H_{2}$ react with excess of $N_{2} = \frac{2}{3} \times 6 = 4 moles$
Hence $4$ moles of ammonia are formed when $6$ moles of $N_{2}$ react with $6$ moles of $H_{2}$ .

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The balanced chemical equation for the formation of ammonia gas by the reaction between nitrogen gas an hydrogen gas is given.N2+3H→2NH3How many moles of ammonia are formed when 6 moles of N2 react with 6 moles of H2?

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