Question

Question #1d7ef

Answer 1

82.5 g of silver oxide decomposed.

Explanation:

You're dealing with the decomposition reaction of silver oxide, `Ag_2O`, into silver metal, `Ag`, and oxygen gas, `O_2`.

The balanced chemical equation for this reaction looks like this

`color(red)(2)Ag_2O_((s)) -> 4Ag_((s)) + O_(2(g))`

You know that you have a `color(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

`PV = nRT => n= (PV)/(RT)`

`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"` `Ag_2O`

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

`0.356cancel("moles") * "23.74 g"/(1cancel("mole")) = color(green)("82.5 g "Ag_2O)`