Zn+2HCL ---> ZnCl2+H2 (g) . How much ZnCl2 can you make from 5 g of Zn?

Answer 1

Balanced equation is Zn(s) + 2 HCl(aq) ==> ZnCl₂(aq) + H₂(g)

moles of Zn present = 5 g x 1 mole/65.38 g = 0.07648 moles

Assuming that HCl is not limiting moles ZnCl₂ that can be produced is 0.07648 moles since balanced equation shows a 1:1 mole ratio.

Mass ZnCl₂= 0.07648 moles x 136.3 g/mole = 10.4 g = 10 g (to 1 significant figure)