Question

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Given 10 g of Zn and 10 g HCl; what volume of Hy is produced?

Answer 1

Answer: 3.02 L of
`H_2` will be produced from the given masses of both reactants.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Zn=(10g)/(65g/mol)=0.15moles`

`\text{Moles of} HCl=(10g)/(36.5g/mol)=0.27moles`

`Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)`

According to stoichiometry :

2 moles of
`HCl` require = 1 mole of
`Zn`

Thus 0.27 moles of
`HCl` will require=
`(1)/(2)* 0.27=0.135moles` of
`Zn`

Thus
`HCl` is the limiting reagent as it limits the formation of product and
`Zn` is the excess reagent.

As 2 moles of
`HCl` give = 1 moles of
`H_2`

Thus 0.27 moles of
`HCl` give =
`(1)/(2)* 0.27=0.135moles` of
`H_2`

Volume of
`H_2=moles* {\text {Molar Volume}}=0.135moles* 22.4g/L=3.02L`

Thus 3.02 L of
`H_2` will be produced from the given masses of both reactants.