Finding mass of lithopone from limiting reactant in BaS and ZnSO4 reaction

[geffman1]

Homework Statement

hey guys I've got a problem, could you tell us where i went wrong

Lithopone is a brilliant white pigment used in water-based interior paints. It is a mixture of barium sulfate and zinc sulfide produced by the reaction
BaS(aq) + ZnSO4(aq) ZnS(s) + BaSO4(s)
What mass of lithopone in grams (to 3 significant figures) is produced in the reaction of 227 mL of 0.257 M ZnSO4 and 325 mL of 0.280 M BaS?

MY WORKED SOLUTION"
Mole of BaS=0.280*0.325=0.091mole
mole of ZnSO4=0.0257*0.277=0.0712moles <therefore this is limiting

i don't know what i should take the molarmass of to work out the mass?

ANSWER IS 1.93 grams

anyhelp would be GREATFULLY appreciated.

Homework Equations

The Attempt at a Solution

 

[Borek]

mole of ZnSO4=0.0257*0.277=0.0712moles <therefore this is limiting

Check your math. Check it twice, as abvious typo is not the error I am thinking about.

ANSWER IS 1.93 grams

No, 1.93g is not the answer. Probably typo again.

You are on the right track. Find limiting reagent, find moles of both solids that precipitated, find their masses - and as you are looking for a precipitate mass, add them.