How many grams of aluminu?m will be produced if 12.0g of Ba reacts with 9.0g of Al2(SO4)3

Question

anonymous · Answer

The reaction gives one type of precipitate, i.e. BaSO4
then, # mole of Ba = 12 / 137.3 = 0.0874 mole
then # of mole of SO4 2- = 9/342.2 (molar mass of Al2(SO4)3) = 0.0263 mole
since mole ratio of Ba: Al2(SO4)3 = 3:1,
therefore, Ba is in excess.
Thus, only 0.0263 mol Ba reacted to give BaSO4
since 3 mole of BaSO4 will be produced, 
so # mole of BaSO4 produced = 3 * 0.0263=0.0789 mole 
gram of BaSO4 = 0.0789 * 233.4 = 18.42g//
correct me if I'm wrong ;)

anonymous · Answer

$$3Ba + Al2(SO4)3 \rightarrow 3BaSO4 + 2Al ^{3+}$$

anonymous · Answer

This is a problem of limiting reactants