Upon mixing `50.0 mL` of `0.1 M` lead nitrate solution with `50.0 mL` of `0.05 M` chromic sulphate solution, precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also, calculate the molar concentration of the species left behind in the final solution. Which is the limiting reagent?

Explanation

`underset({:(3 mmol),("Initial" 50 xx 0.1),(= 5 mmol):})(3 Pb(NO_(3))_(2)) + underset({:(1 mmol),(50 xx 0.05),(2.5 mmol):})(Cr_(2) (SO_(4))_(3)) rarr underset(3 mmol)(3 PbSO_(4) darr) + underset(2 mmol)(Cr(NO_(3))_(3))`
a. 3 mmol of `Pb(NO_(3))_(2) = 1` mmol of `Cr(SO_(4))_(3)`
5 mmol of `Pb(NO_(3))_(2) = (1)/(3) xx 5 = 1.66 mmol`
b. Since `Pb(NO_(3))_(2)` is completely, so it is a limiting reagent.
`Cr_(2) (SO_(4))_(3) = 2.5 - 1.66 = 0.833 mmol`
3 mmol of `Pb (NO_(3))_(2) = 3 mmol of PbSO_(4)`
5 mmol of `Pb(NO_(3))_(2) = 5 mmol of PbSO_(4)`
`= 5 xx 10^(-3) mol`
`0.005 mol of PbSO_(4)`
c. 3 mmol of `Pb(NO_(3))_(2) = 2` mmol of `Cr(NO_(3))_(2)`
5 mmol of `Pb(NO_(3))_(2) = (2)/(3) xx 5 = 3.33 mmol`
Total volume `= 50 + 50 = 100 mL`
Concentration of `Cr(NO_(3))_(3)` in the solution
`(mmol)/(mL) = (3.33)/(100) = 0.033 M`
iv. Concentration of `Cr_(2) (SO_(4))_(3)` left `= (0.833)/(100) = 0.0083 M`