Question

For the following balanced equation below determine how many mL of 0.450 M Ba(NO₃)₂ would be needed to react with 14.8 g of Al₂(SO₄)₃

  

Al₂(SO₄)₃(s)   +   3Ba(NO₃)₂(aq) ---> 3BaSO₄(s) + 2Al(NO­₃)₃(aq)

Please show all work

Answer 1

The given reaction is already balanced interms of coefficients

Al₂(SO₄)₃(s)   +   3Ba(NO₃)₂(aq) ---> 3BaSO₄(s) + 2Al(NO­₃)₃(aq)

1 mol                    3 mol                    3 mol             2 mol

1 mol Al₂(SO₄)₃ requires 3 mol Ba(NO₃)₃

1) first we will calculate how many moles present in 14.8 g of Al₂(SO₄)₃

moles = mass of Al₂(SO₄)₃ in g / molar mass of Al₂(SO₄)₃ in g/mol

          = 14.8 g / 342.15 g/mol

          = 0.04325588192 mol

2) 1 mol Al₂(SO₄)₃ requires 3 mol Ba(NO₃)₃

    0.04325588192 mol Al₂(SO₄)₃ requires 3 x 0.04325588192 = 0.12976764576 mol Ba(NO₃)₃

3) suppose y mL x 10^-3 L x 0.450 M = 0.12976764576 mol

y mL = 288.37 mL

288.37 mL of 0.450 M Ba(NO₃)₂ would be needed to react with 14.8 g of Al₂(SO₄)₃

Answer = 288.37 mL

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