Question

In this reaction, 4.58L of `O_2` were formed at 745 mmHg and 308K. How many grams of `Ag_2O` decomposed? `2Ag_2O_((s)) -> 4Ag_((s)) + O_(2(g))`

Answer 1

Your reaction required the decomposition of 82.5 g of silver oxide.

Start with the balanced chemical equation

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

Notice that you have a `color(red)(2):1` mole ratio between silver oxide and oxygen gas. This tells you that, regardless of how many moles of oxygen the reaction produced, two times more moles of silver oxide underwent decomposition.

Use the ideal gas law equation to determine how many moles of oxygen were produced

`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`

Use the aforementioned mole ratio to determine how many moles of silver oxide reacted

`0.178cancel("moles"O_2) * (color(red)(2)" moles"Ag_2O)/(1cancel("mole"O_2)) = "0.356 moles"` `Ag_2O`

Now use silver oxide's molar mass to determine how many grams would contain this many moles

`0.356cancel("g") * ("1 mole" Ag_2O)/(231.74cancel("g")) = color(green)("82.5 g")`