Answer to Question #16675 in Inorganic Chemistry for Zach
If 25.0 grams of Barium Sulfate are produced during the reaction, how many grams of Iron(III) sulfate were used?
& x & 25.0
3 BaCl2 + Fe2(SO4)3 ---> 3 BaSO4 + 2 FeCl3
& 400 3* 233
x=400*25 / (3*233)=14.306 g
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