Question

Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl 2 ( aq ) + H 2 ( g ) How many milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn ( s ) ?

Answer 1

Answer: 12.0 milliliters of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn.

Step-by-step explanation:

moles =
`\frac{\text {given mass}}{\text {Molar mass}}`

moles of zinc =
`(2.55g)/(65.38g/mol)=0.0390moles`

The balanced chemical equation is :

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

According to stoichiometry:

1 mole of zinc reacts with = 2 moles of HCl

Thus 0.0390 moles of zinc reacts with =
`(2)/(1)* 0.0390=0.0780` moles of HCl

To calculate the volume for given molarity, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution in ml}}` .....(1)

Molarity of
`HCl` solution = 6.50 M

Volume of solution = ?

Putting values in equation 1, we get:

`6.50M=\frac{0.0780* 1000}{\text{Volume of solution in ml}}`

`{\text{Volume of solution in ml}}=12.0ml`

Thus 12.0 ml of 6.50 M HCl ( aq ) are required to react with 2.55 g Zn