# Question 1d7ef

Jun 20, 2015

82.5 g of silver oxide decomposed.

#### Explanation:

You're dealing with the decomposition reaction of silver oxide, $A {g}_{2} O$, into silver metal, $A g$, and oxygen gas, ${O}_{2}$.

The balanced chemical equation for this reaction looks like this

$\textcolor{red}{2} A {g}_{2} {O}_{\left(s\right)} \to 4 A {g}_{\left(s\right)} + {O}_{2 \left(g\right)}$

You know that you have a $\textcolor{red}{2} : 1$ mole ratio between silver oxide and oxygen gas. This means that, regardless of how many moles of oxygen gas the reaction produced, you had 2 times more moles of silver oxide that underwent decomposition.

In order to determine how many moles of oxygen gas the reaction produced, use the ideal gas law equation

$P V = n R T \implies n = \frac{P V}{R T}$

n_(O_2) = (745/760cancel("atm") * 4.58cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * 308cancel("K")) = "0.178 moles" ${O}_{2}$

Notice that I converted the pressure from mmHg to atm.

If this many moles of oxygen were produced, then the number of moles of silver oxide that underwent decomposition was

0.178cancel("moles"O_2) * (color(red)(2)" moles "Ag_2O)/(1cancel("mole"O_2)) = "0.356 moles"# $A {g}_{2} O$

To determine the mass in grams of silver oxide that contains this many moles, use the compound's molar mass

$0.356 \cancel{\text{moles") * "23.74 g"/(1cancel("mole")) = color(green)("82.5 g } A {g}_{2} O}$