A 77.0-mL sample of a 0.203 M lithium sulfate solution is mixed with 55.0 mL of...

A 77.0-mL sample of a 0.203 M lithium sulfate solution is mixed with 55.0 mL of a 0.226 M lead(II) nitrate solution and this precipitation reaction occurs:

Li₂SO₄(aq) + Pb(NO₃)₂(aq) ⟶ 2 LiNO₃(aq) + PbSO₄(s)

The solid PbSO₄ is collected, dried, and found to have a mass of 4.71 g. Determine the percent yield. 1 mole PbSO₄ = 303.26 g

Group of answer choices

A.) 64.7 %

B.) 71.9 %

C. 98.5 %

D.) 80.0 %

Solutions

Expert Solution

queen_honey_blossom answered 1 year ago

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