When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, _________ \(\times\) \(10^{-5}\) moles of lead sulphate precipitate out. (Round off to the Nearest Integer).
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Correct Answer: 525
Solution and Explanation
Moles of Pb(NO3)2 = 35 mL \(\times\) 0.15 M = 5.25 mmol.
Moles of Cr2(SO4)3 = 20 mL \(\times\) 0.12 M = 2.4 mmol.
Cr2(SO4)3 needed for 5.25 mmol Pb(NO3)2 = 5.25/3 = 1.75 mmol, which is less than 2.4 mmol, so Pb(NO3)2 is the limiting reagent.
Moles of PbSO4 = 5.25 mmol = \(5.25\times 10^{-3}\) mol = \(525\times 10^{-5}\) mol.
Answer: 525
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