Question

How many moles of N2O5 are needed to produce 7.90 g of NO2? 2N2O5 = 4NO2 + O2

Answer 1

moles of N2O5 = moles of NO2 * ( 2 moles of N2O5 / 4 moles of NO2

Answer 2

Answer: 0.08585 moles of $N_2O_5$ is needed to produce 7.9 grams of $NO_2$

Explanation: Moles is calculated by using the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of $NO_2$ = 46 g/mol

Number of moles of $NO_2=\frac{7.9g}{46g/mol}$ = 0.1717 moles

For the given chemical reaction:

$2N_2O_5\rightarrow 4NO_2+O_2$

By Stoichiometry,

4 moles of $NO_2$ is produced by 2 moles of $N_2O_5$

So, 0.1717 moles of $NO_2$ is produced by = $\frac{2}{4}\times 0.1717$ = 0.08585 moles of $N_2O_5$

Hence, moles of $N_2O_5$ is 0.08585 moles.