Question

How many moles of ZnCl2 will be produced from 61.0 g of Zn, assuming HCl is excess?

Answer 1

Answer: 0.938 moles of
`ZnCl_2` will be produced.

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

moles of zinc:

`\text{Number of moles}=(61.0g)/(65g/mol)=0.938moles`

`Zn+2HCl\rightarrow ZnCl_2+H_2`

As HCl is in excess , zinc is the limiting reagent and it limits the formation of product.

According to stoichiometry :

1 mole of Zn produce = 1 mole of
`ZnCl_2`

Thus moles of Zn produce =
`(1)/(1)* 0.938=0.938` moles of
`ZnCl_2`

Thus 0.938 moles of
`ZnCl_2` will be produced.