Question

How many moles of NO₂ would be produced from 4.5 mol of O₂ in the reaction below assuming the reaction has a 57.0% yield?

2 NO (g) + O₂ (g) → 2 NO₂ (g)

Answer 1

From the reaction,

$2 NO (g) + O₂ (g) → 2 NO₂ (g)$

One mole of $O_2$ produces two moles of $NO_2$ theoretically .

So, theoretical yield$(Y_t)$ of $4.5$ moles of $O_2$ will be $4.5\times 2= 9$ moles of $NO_2$ .

Let $Y_a$ Be the actual yield of $NO_2.$

Percentage yield $= \frac{Y_a}{Y_t} \times 100$

Substituting the values,

$57= \frac{Y_a}{9} \times 100$

$\implies Y_a= \frac{57\times 9}{100}=5.13$

So, $5.13$ moles of $NO_2$ will pe produced from $4.5$ moles of $O_2.$