Question

what quantity, in moles, of no₂ would be required to produce 4.95 mol of hno₃ in the presence of excess water in the following chemical reaction? 3no₂(g) + h₂o(l) ⇌ 2hno₃(g) + no

Explanation:

Step1: Identify mole - ratio

From the balanced equation $3NO_2(g)+H_2O(l)
ightleftharpoons 2HNO_3(g)+NO(g)$, the mole - ratio of $NO_2$ to $HNO_3$ is $3:2$.

Step2: Set up proportion

Let $x$ be the moles of $NO_2$ required. We have the proportion $\frac{x}{4.95\ mol\ HNO_3}=\frac{3\ mol\ NO_2}{2\ mol\ HNO_3}$.

Step3: Solve for $x$

$x=\frac{3\ mol\ NO_2}{2\ mol\ HNO_3}\times4.95\ mol\ HNO_3$.
$x = 7.425\ mol$.

Answer:

$7.425$ mol