Question

How many moles of ZnCl 2 will be produces from 21.0 g of Zn, assuming CuCl 2 is
available in excess?
ZN+CnCl2+Cu

Answer 1

$CuCl_2+Zn→ZnCl_2+Cu$

1 mole of$Zn$produce 1 mole of $ZnCl_2$

$1\:mole\:Zn=1\:mole\:ZnCl_2$

Atomic mass of Zn =65

Given massof Zn = 21

$n_{moles} =\frac{given\:mass}{atomic\:mass}\\\\n_{moles}Zn=21/65=0.32$

$1\:mole\:of\:Zn→1\:mole\:of\:ZnCl_2\\0.32\:mole\:of\:Zn→0.32\:mole\:of\:ZnCl_2$

0.32 moles of $ZnCl_2$ will be formed