# How many Oxygen atoms in 1.44 x 10^5 kg of Nitrogen Oxide?

Jul 17, 2017

Well, I will do it for $N {O}_{2} \left(g\right)$........and get over six million moles......

#### Explanation:

But of course there is $N O$ and ${N}_{2} O$.........

For $N {O}_{2}$, we have a molecular mass of $46.01 \cdot g \cdot m o {l}^{-} 1$.

And thus $\text{Moles of}$ $N {O}_{2}$ $=$ $\frac{1.44 \times {10}^{8} \cdot g}{46.01 \cdot g \cdot m o {l}^{-} 1}$

$\equiv 3.13 \times {10}^{6} \cdot m o l$....and thus $6.26 \times {10}^{6} \cdot m o l$ $\text{oxygen atoms.}$

Then, you continue to solve for the number of oxygen atoms.

$6.26 \times {10}^{6} \cdot m o l \times 6.022 \cdot {10}^{23} \cdot m o {l}^{-} 1 \cong 4.0 \times {10}^{30}$ $\text{oxygen atoms}$ which is nearer to the given answer.