Saved Bookmarks
| 1. |
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? |
|
Answer» `{:(,3Pb(NO_(3))_(2)+,Cr_(2)(SO_(4))_(3) ,rarr,3PbSO_(4)darr+,2Cr(NO_(3))_(3),,,),("Meq.",45 xx 0.25xx2,25 xx 0.1 xx 6,,,,,,),("before",=22.5,=15,,0,0,,,),("reaction",,,,,,,,),("Meq. after",7.5,0,,15,15,,,),("reaction",,,,,,,,):}` `:.` Meq. Of `PbSO_(4)` precipitated `= 15` `:.` millie-mole of `PbSO_(4)` precipitated `= (15)/(2)` `:.` Mole of `PbSO_(4)` precipitated `= (15)/(2) xx (1)/(1000) = 0.0075` `because [ ] = ("Meq.")/("total volume" xx "valency" )` Also `[Pb^(2+)] = (7.5)/(70 xx 2) = 0.0536 M` `[NO_(3)^(-)] = (7.5 +15)/(70 xx 1) = 0.32 M` `[Cr^(3+)] = (15)/(70 xx 3) = 0.0714 M` |
|