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Question

Exactly 500.0 mL of a solution containing 316.2 ppm of Ba(NO3)2 is mixed with 150.0 mL of 0.047 M Al2(SO4)3 solution. (N: 14.0 ; O:16.0; S:32.0; Al:27; Ba:137 g/mol)

What mass of solid BaSO4 is formed?

What is the molar concentration of the unreacted reagent?

please write the solution with steps

Answer 1

3Ba(NO₃)₂ + Al₂(SO₄)₃ ==> 3BaSO₄(s) + 2Al(NO₃)₃ ... balanced equation

To find the mass of BaSO₄ formed, we first must find which reactant is limiting:

moles Ba(NO₃)₂ = 500.0 ml x 1 L/1000 ml x 316.2 mg/L x 1 g/1000 mg x 1 mol/261 g = 6.057x10^-4 moles

moles Al₂(SO₄)₃ = 150.0 ml x 1 L/1000 ml x 0.047 mol/L = 7.05x10^-3 moles

So, Ba(NO₃)₂ is limiting and determines moles of BaSO₄ that will be produced

mass BaSO₄ produced = 6.057x10^-4 mol Ba(NO₃)₂ x 3 mol BaSO₄ / 3 mol Ba(NO₃)₂ x 233 g/mol = 0.1441 g

Molar concentration of unreacted reagent would refer to Al₂(SO₄)₃:

moles left = 7.05x10^-3 mol - 0.6057x10^-3 mol = 6.44x10^-3 moles / 0.650 L = 9.91 M