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?

Jan 28, 2017

Approx. $60 \cdot g$ of ${K}_{2} S {O}_{4}$ were initially used.

Explanation:

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

$2 A l B {r}_{3} \left(a q\right) + 3 {K}_{2} S {O}_{4} \left(a q\right) \rightarrow A {l}_{2} {\left(S {O}_{4}\right)}_{3} \left(s\right) \downarrow + 6 K C l \left(a q\right)$

Molar quantities are as follows:

$A l B {r}_{3} ,$ (?*g)/(133.35*g*mol^-1)

${K}_{2} S {O}_{4} ,$ (?*g)/(174.26*g*mol^-1)

$A {l}_{2} {\left(S {O}_{4}\right)}_{3} ,$ $\frac{37.5 \cdot g}{342.15 \cdot g \cdot m o {l}^{-} 1} = 0.110 \cdot m o l$

Given the stoichiometry, initially there were $2 \times 0.110 \cdot m o l$ $A l B {r}_{3}$, and $3 \times 0.110 \cdot m o l$ ${K}_{2} S {O}_{4}$. Do you appreciate this? If no, say the word, and we will try again.

Given $3 \times 0.110 \cdot m o l$ ${K}_{2} S {O}_{4}$, 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.