Question

With 21 g of Zinc, and 7 g of CuCl2, how much ZnCl2 is made in grams?

Answer 1

Answer: 7.07 grams

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$    

$\text{Moles of} zinc=\frac{21g}{65g/mol}=0.32moles$

$\text{Moles of} CuCl_2=\frac{7g}{134g/mol}=0.052moles$

$Zn+CuCl_2\rightarrow Cu+ZnCl_2$

According to stoichiometry :

1 mole of $CuCl_2$ require 1 mole of $Zn$

Thus 0.052 moles of $CuCl_2$ will require=$\frac{1}{1}\times 0.052=0.052moles$  of $Zn$

Thus $CuCl_2$ is the limiting reagent as it limits the formation of product and $Zn$ is the excess reagent.

As 1 mole of $CuCl_2$ give = 1 mole of $ZnCl_2$

Thus 0.052 moles of $CuCl_2$ give =$\frac{1}{1}\times 0.052=0.052moles$  of $ZnCl_2$

Mass of $ZnCl_2=moles\times {\text {Molar mass}}=0.052moles\times 136g/mol=7.07g$

Thus 7.07 g of $ZnCl_2$ will be produced from the given masses of both reactants.