Upon mixing 50.0 mL of 0.1 M lead nitrate solution with 50 mL of 0.05 M chromic sulphate solution, precipitation of lead sulphate solution takes place. How many moles of lead sulphate are formed and what is the molar concentration of chromic sulphate left in the solution?

3 mmol           1 mmol            3 mmol         2mmol3PbNO32+Cr2SO43→3PbSO4↓+2CrNO3350×0.1                                                       50 ×0.05       =5 mmol                                                      =2.5 mmol First find the limiting reagent. 3 mmol of PbNO32⇒1 mmol of Cr2SO435 mmol of PbNO32⇒13×5⇒53=1.66mmol So PbNO32 is the limiting reagent. i.  3mmol of PbNO32⇒3 mmol of PbSO45 mmol of PbNO32⇒5 mmol⇒51000mol⇒0.005 molii. Species left in the solution are Cr2SO43 and CrNO33.  To calculate the concentration of Cr2SO43 :  Initial mmol=2.5 Reacted mmol=1.65 Left mmoles =2.5−1.66=0.84mmol Total Volume =50+50=100mL Concentration =0.84100=0.0084Miii. To calculate the concentration of CrNO333 m mol of Pb(NO3)2 ⇒23×5⇒103=3.33 mmolConcentration = 3.33100=0.033M