Answer to Question #67043 in General Chemistry for johnson

Question

what mass of ZnCl2 can be prepared from the reaction of 1.69 of zinc with 1.10 grams of HCL

Answer 1

Zn+2HCl->ZnCl₂+H₂

One mole Zn reacts with 2 mole HCl

V(Zn)=m/M=1.69g/65g/mol=0.025mol

V(HCl)=m/M=1.10g/36.5g/mol=0.030Mole

0.025mole(Zn):0.030mole(HCl)=1:1.12

Zn - in excess, HCl - In want

2mole(HCl)/0.025mole(HCl)=1mole(ZnCl₂)/xmole(ZnCl₂)

V(ZnCl₂)=0.0125mole

m(ZnCl₂)=V*M=0.0125mole*136g/mole=1.7g