When 35 mL of 0 . 15 M lead nitrate solution is mixed with 20 mL of 0 . 12 M chromic sulphate solution, ____________ × 10 - 5 moles of lead sulphate precipitate out. (Round off to the Nearest Integer).

Step-by-step solution

3   Pb NO 3 2 + Cr 2 SO 4 3 → 3 PbSO 4 + 2 Cr NO 3 3 35   ml                 20   ml 0 . 15   M 0 . 12   M = 5 . 25   m . mol = 2 . 4   m . mol 5 . 25   m . mol = 5 . 25 × 10 - 3   mol Therefore moles of PbSO 4 formed = 5 . 25 × 10 - 3 = 525 × 10 - 5