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.

Explanation:

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

moles of zinc =$\frac{2.55g}{65.38g/mol}=0.0390moles$

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 = $\frac{2}{1}\times 0.0390=0.0780$ moles of HCl

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

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

Molarity of $HCl$ solution = 6.50 M

Volume of solution = ?

Putting values in equation 1, we get:

$6.50M=\frac{0.0780\times 1000}{\text{Volume of solution in ml}}$

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

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