Question

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

Answer 1

Answer: 7.07 grams

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}*{\text{Molar Mass}}`

`\text{Moles of} zinc=(21g)/(65g/mol)=0.32moles`

`\text{Moles of} CuCl_2=(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=
`(1)/(1)* 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 =
`(1)/(1)* 0.052=0.052moles` of
`ZnCl_2`

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

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