# Question #5cb6e

Jun 8, 2017

We need (i) a stoichiometric equation:

$N O \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow N {O}_{2} \left(g\right)$ and gets approx. $1 \cdot g$ $N {O}_{2}$.

#### Explanation:

And (ii) we work out equivalent quantities of $N O \left(g\right)$ and $\text{dioxygen gas}$......

$\text{Moles of NO} = \frac{0.66 \cdot g}{30.01 \cdot g \cdot m o {l}^{-} 1} = 0.0212 \cdot m o l .$

$\text{Moles of dioxygen} = \frac{0.58 \cdot g}{32.00 \cdot g \cdot m o {l}^{-} 1} = 0.0181 \cdot m o l .$

Given the stoichiometric equation, dioxygen is in EXCESS, and given quantitative reaction (which is not at all likely), we thus make $0.0212 \cdot m o l$ of $N {O}_{2} \left(g\right)$, a mass of $0.0212 \cdot m o l \times 46.01 \cdot g \cdot m o {l}^{-} 1 = 0.98 \cdot g$.