If 95.00 mL of 0.2500 M barium nitrate are combined with 150.00 mL of 0.1000 M lithium sulfate how many barium ions are in the resulting solution in units of millimoles ?

Answer 1

Barium nitrate = Ba(NO₃)₂

Lithium sulfate = Li₂SO₄

Reaction: Ba(NO₃)₂(aq) + Li₂SO₄(aq) ==> BaSO₄(s) + 2LiNO₃(aq)

moles Ba(NO₃)₂ present = 0.095 L x 0.2500 mol/L = 0.02375 mol Ba(NO₃)₂

moles Li₂SO₄ present = 0.1500 L x 0.1000 mol/L = 0.01500 mol Li₂SO₄

Limiting reactant = Li₂SO₄

moles Ba(NO₃)₂ used in the reaction = 0.01500 (since 1 mol Li₂SO₄ reacts with 1 mol Ba(NO₃)₂

moles Ba(NO₃)₂ left over = 0.02375 mol - 0.01500 mol = 0.00875 mole Ba(NO₃)₂ left over

moles Ba²⁺ left over = 0.00875 moles

Final volume = 95 ml + 150 ml = 245 ml = 0.245 L

Final [Ba^2+] = 0.00875 mol/0.245 L = 0.03571 M