Question

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N2 + 3H2 >> 2NH3 If the reaction yield is 87.5% , how many moles of N2 are needed to produce 3.00 mol of NH3 ?

Answer 1

Answer on Question #55234 – Chemistry – General Chemistry

Question:

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: $N_2 + 3H_2 \gg 2NH_3$. If the reaction yield is 87.5%, how many moles of $N_2$ are needed to produce 3.00 mol of NH₃?

Solution:

v – The number of moles (mol);

$N_2 + 3H_2 \rightarrow 2NH_3;$

$v(NH_3) = 3.00\ \text{mol};$

$\mu = 87.5\%$

Calculate the theoretical yield of the reaction:

$v(NH_3\ \text{theoretical}) = v(NH_3 - \text{obtained during the reaction}) / \mu;$

$v(NH_3\ \text{theoretical}) = \frac{v(NH_3)}{\mu}; \quad v(NH_3\ \text{theoretical}) = \frac{3}{0.875} = 3.423\ \text{mol};$

$N_2 + 3H_2 \rightarrow 2NH_3$; According to the equation: $v(NH_3\ \text{theoretical}) : v(N_2) = 2 : 1;$

$v(N_2) = \frac{v(NH_3\ \text{theoretical})}{2} = \frac{3.423}{2} = 1.712\ \text{mol};$

$v(N_2) = 1.712\ \text{mol};$

Answer: 1.712 mol.

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