# How many moles "Ag"_2"O" are needed to produce "4.24 mol O"_2"?

## $\text{2Ag"_2"O}$$\rightarrow$$\text{4Ag + O"_2}$

May 21, 2016

$\text{8.50 mol Ag"_2"O}$ are needed to produce $\text{4.25 mol O"_2}$.

#### Explanation:

Balanced Equation

$\text{2Ag"_2"O}$$\rightarrow$$\text{4Ag"+"O"_2}$

Multiply the given moles of $\text{O"_2}$ times the mole ratio between $\text{Ag"_2"O}$ and $\text{O"_2}$ from the balanced equation, with $\text{Ag"_2"O}$ as the numerator.

$4.25 \cancel{\text{mol O"_2xx(2"mol Ag"_2"O")/(1cancel"mol O"_2)="8.50 mol Ag"_2"O}}$