Question

Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g) How many milliliters of 2.50 M HCl(aq) are required to react with 4.65 g of an ore containing 50.0% Zn(s) by mass? volume: mL

Answer 1

Volume of HCl required = 28.4 mL

Step-by-step explanation:

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

Mass of zinc ore = 4.65 g

% of zinc in zinc ore = 50 %

So, mass of zinc in zinc ore = 0.50 × 4.65 = 2.325 g

No. of moles of Zn =
`(2.325)/(65.40) = 0.0355\ mol`

So, as per the reaction coefficient,

1 mol of zinc reacts with 2 mol of HCl

0.0355 mol of zinc reacts with

= 0.0355 × 2 = 0.071 mol of HCl

Molarity of HCl = 2.50 M

Volume of HCl =
`(Moles)/(Concentration)`

`Volume\ of\ HCl = (0.071)/(2.50) = 0.0284\ L`

1 L = 1000 mL

0.0284 L = 1000 × 0.0284 = 28.4 mL