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?

1 Answer
Jan 28, 2017

Answer:

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.