Asked by Anonymous
Suppose 500 ml of a .0035 M solution of lead nitrate is reacted with sulfuric acid. What mass of lead sulfate should precipitate?
Answers
Answered by
DrBob222
Pb(NO3)2 + H2SO4 ==> PbSO4 + 2HNO3
mols Pb(NO3)2 = M x L = ?
mols PbSO4 = moles Pb(NO3)2 from the equation which is 1:1.
mols PbSO4 x molar mass = grams.
mols Pb(NO3)2 = M x L = ?
mols PbSO4 = moles Pb(NO3)2 from the equation which is 1:1.
mols PbSO4 x molar mass = grams.
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