Question

the theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to product. consider the reaction: n₂(g) + o₂(g) → 2 no(g). if 19.50 g n₂ is mixed with 5.630 g o₂, calculate the theoretical yield (g) of no produced by the reaction.

Explanation:

Step1: Calculate moles of reactants

The molar mass of $N_2$ is $M_{N_2}=28.02\ g/mol$, and for $O_2$ is $M_{O_2} = 32.00\ g/mol$. The number of moles of $N_2$, $n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{19.50\ g}{28.02\ g/mol}\approx0.696\ mol$. The number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{5.630\ g}{32.00\ g/mol}\approx0.176\ mol$.

Step2: Determine the limiting reactant

From the balanced - chemical equation $N_2(g)+O_2(g)
ightarrow2NO(g)$, the mole ratio of $N_2$ to $O_2$ is 1:1. Since $n_{O_2}=0.176\ mol$ and $n_{N_2}=0.696\ mol$, $O_2$ is the limiting reactant because it will be completely consumed first.

Step3: Calculate moles of NO produced

The mole ratio of $O_2$ to $NO$ is 1:2. So the number of moles of $NO$ produced, $n_{NO}=2\times n_{O_2}$. Since $n_{O_2} = 0.176\ mol$, $n_{NO}=2\times0.176\ mol = 0.352\ mol$.

Step4: Calculate the mass of NO produced

The molar mass of $NO$ is $M_{NO}=14.01\ g/mol + 16.00\ g/mol=30.01\ g/mol$. The mass of $NO$ produced, $m_{NO}=n_{NO}\times M_{NO}=0.352\ mol\times30.01\ g/mol\approx10.57\ g$.

Answer:

$10.57\ g$