Question

how many grams of precipitate could i make if i started with 6.00mL of 0.15M silver nitrate and 5.00 mL of 0.10M strontium chloride

Answer 1

SrCl2 + 2AgNO3 -----> 2AgCl + Sr(NO3)2

no of moles of AgNO3 = molarity * volume in L

                                  = 0.15*0.006

                                    = 0.0009 moles of AgNO3

no of moles of SrCl2      = Molarity * volume in L

                                   = 0.1*0.005 = 0.0005 moles of SrCl2

limiting reagent is SrCl2

1 mole of SrCl2 react with AgNO3 to from 2 moles of AgCl

0.0005 moles of SrCl2 react with AgNo3 to from = 2*0.0005/1 = 0.001 mole of AgCl

mass of AgCl = no of moles * molar mass

                   = 0.001*143.4 = 0.1434gm of AgCl