Question

what is the mass of reagent of 10g of zn and 10g of HCl after determining the LR. Im having trouble calculating the mass of the excess Zn

Answer 1

     Zn + 2HCl --------------> ZnCl2 + H2

no of moles of Zn = W/G.A.Wt

                             = 10/65.4   = 0.153 moles

no of moles of HCl   = W/G.M.Wt

                                = 10/36.5   = 0.274 moles

1 mole of Zn react with 2 moles of HCl

0.153 moles of Zn react with = 2*0.153/1   = 0.306 moles of HCl

HCl is limiting reactant

Zn + 2HCl --------------> ZnCl2 + H2

2 moles of HCl react with 1 mole of Zn

0.274 moles of HCl react with = 1*0.274/2   = 0.137 moles of Zn

The no of moles of excess reactant left after complete the reaction = 0.153-0.137   = 0.016 moles of Zn

mass of Zn   = no of moles * gram atomic mass

                    = 0.016*65.4   = 1.0464 g >>>>answer