anonymous
  • anonymous
How many grams of aluminu?m will be produced if 12.0g of Ba reacts with 9.0g of Al2(SO4)3
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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 ;)
anonymous
  • anonymous
\[3Ba + Al2(SO4)3 \rightarrow 3BaSO4 + 2Al ^{3+}\]
anonymous
  • anonymous
This is a problem of limiting reactants

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