Question

A `37.5*g` mass of `"aluminum sulfate"` was obtained from the reaction between `AlBr_3` and `"potassium sulfate"`. What were the starting masses of the reactants?

Answer 1

Approx. `60*g` of `K_2SO_4` were initially used.

Explanation:

We follow the stoichiometric equation (which you have represented):

`2AlBr_3(aq) + 3K_2SO_4(aq) rarr Al_2(SO_4)_3(s)darr + 6KCl(aq)`

Molar quantities are as follows:

`AlBr_3,` `(?*g)/(133.35*g*mol^-1)`

`K_2SO_4,` `(?*g)/(174.26*g*mol^-1)`

`Al_2(SO_4)_3,` `(37.5*g)/(342.15*g*mol^-1)=0.110 *mol`

Given the stoichiometry, initially there were `2xx0.110*mol` `AlBr_3`, and `3xx0.110*mol` `K_2SO_4`. Do you appreciate this? If no, say the word, and we will try again.

Given `3xx0.110*mol` `K_2SO_4`, this constitutes a mass of `0.330*molxx174.26*g*mol^-1=??`

Note that the question is a bit naive and does not reflect chemical reality. As I recall, aluminum sulfate has some solubility in aqueous solution, and the assumption that it would all precipitate is incorrect.