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

1. anonymous

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 ;)

2. anonymous

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

3. anonymous

This is a problem of limiting reactants