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30 July, 17:24

If 100.0 mL of 0.105 M Na 2 SO 4 are added to 100.0 mL of 0.985 M Pb (NO 3) 2, how many grams of PbSO 4 can be produced? Na 2 SO 4 (aq) + Pb (NO 3) 2 (aq) ⟶ 2 NaNO 3 (aq) + PbSO 4 (s)

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  1. 30 July, 17:29
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    3.18 g of lead sulfate is produced in this reaction

    Explanation:

    The reaction is this one:

    Na₂SO₄ (aq) + Pb (NO₃) ₂ (aq) → 2 NaNO₃ (aq) + PbSO₄ (s)

    Let's determine the moles of reactants, to find out which is the limiting.

    Molarity. volume = Mol

    0.105 m/L. 0.100L = 0.0105 moles of sulfate

    0.985 m/L. 0.100L = 0.0985 moles of nitrate

    Ratio is 1:1, so 1 mol of sulfate needs 1 mol of nitrate, to react.

    0.0985 moles of nitrate need 0.0985 moles of sulfate to react, but I only have 0.0105 moles so the sulfate is my limiting reactant.

    Ratio is again 1:1, so 0.0105 moles of sulfate make 0.0105 moles of lead sulfate.

    Mol. molar mass = mass → 0.0105 mol. 303.26 g/m = 3.18 g
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