Question

Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 3.50M HCl(aq) are required to react with 2.65g of an ore containing 41.0% Zn(s) by mas

Answer 1

Solution:

The reaction of Zn with HCl is given as,

Zn (s) + HCl (aq) = ZnCl2 (aq) + H2(g)

Since, ore contains 41% of Zn

Hence, weight of Zn in 2.65 g of ore

= 2.65 g x 41 / 100 = 1.0865g

Molar mass of Zn = 65.39 g mol-1

Then, number of moles of Zn = 1.0865 g / 65.39 g mol-1 =0.0166 mol

Since, 1 mol of Zn required 2 mol of HCl to form ZnCl2, therefore,

Number of moles of HCl = 2 x 0.0166 =0.0332 mol

Molarity (M) = number of mole / Volume in L

3.50 M HCl = 0.0332 mol  / Volme of HCl in L

Volume of HCl = 0.0332 mol / 3.50 mol L -1 = 0.009485 L

Volume of HCl in mL = 0.009485 x 1000 =9.485 mL