Question

In this reaction, `4.58 L` of `O_2` were formed at `745 mm Hg` and `308 K`. How many grams of `Ag_2O` decomposed? `2 Ag_2O(s)` → `4 Ag(s)` + `O_2(g)`

Answer 1

We use the Ideal Gas equation to interrogate the number of moles of dioxygen formed...we get approx. of starting mass of `100*g` of the oxide.

Explanation:

`PV=nRT`....and so `n=(PV)/(RT)`...

And you must simply know that `1*atm-=760*mm*Hg`...i.e. that `1*atm` will support a column of mercury that is `760*mm` high...and so here we may use a unit of length to represent a unit of pressure: `760*mm*Hg-=1*atm`....the units of the Universal Gas constant, are `R=0.0821*L*atm*K^-1*mol^-1`

`n=((745*mm*Hg)/(760*mm*Hg*atm)xx4.58*L)/(0.0821*(L*atm)/(K*mol)xx308*K)=0.178*1/(1/(mol))=0.178*mol`.

Now this is a reasonable volume, because we know that the molar volume of an Ideal Gas at standard pressure is around `25*L`. Both the molar volume, and various values of `R` will be quoted to you in an exam....you still have to be able to use them effectively...

And so we got `0.178*mol` dioxygen gas. And given the stoichiometric equation, the which was given to you...

`Ag_2O(s) rarr2Ag(s)+1/2O_2(g)uarr`

We KNOW that `1/2*"equiv"` dioxygen gas were formed with respect to each equiv of silver oxide....and so there were `2xx0.178*mol` with respect to `Ag_2O`..i.e. a mass of...

`2xx0.178*molxx231.74*g*mol^-1~=100*g`....i.e. a large mass of silver oxide .. whew .. arithmetic.