Upon mixing 45.0 mL of 0.25 M lead nitrate solution with 25.0 mL of 0.10 M chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the molar concentrations of species left behind in the final solution. Assume that lead sulphate is completely insoluble.

Explanation

Apply the concept of limiting reagent
`PbSO_4 = 0.0075` mole
`[Pb^(2+)] = 0.05357 M, [NO_3] = 0.3214M`
`[Cr^(3+)] = 0.0714M`