What mass of lead(II) chromate, can be produced by the addition of excess sodium chromate to 25 ml of 0.493 M solution of lead (II) nitrate?

Na₂CrO₄(aq) + Pb(NO₃)₂(aq) → PbCrO₄(s) + 2NaNO₃(aq)
OR: Mole ratio Pb(NO₃)₂ : PbCrO₄ = 1 : 1

No. of moles of Pb(NO₃)₂ reacted = (0.493 mol/L) × (25/1000 L) = 0.0123 mol
No. of moles of PbCrO₄ formed = 0.0123 mol

Molar mass of PbCrO₄ = (207.2 + 52.0 + 16.0×4) g/mol = 323.2 g/mol
Mass of PbCrO₄ formed = (323.2 g/mol) × (0.0123 mol) = 3.98 g

Pb(NO3)2 + Na2CrO4 → PbCrO4 + 2 NaNO3

(0.025 L) x (0.493 mol/L Pb(NO3)2) x (1 mol PbCrO4 / 1 mol Pb(NO3)2) x (323.1937 g PbCrO4/mol) =
4.0 g PbCrO4