Question

Zinc reacts with hydrochloric acid according to the reaction equation

Zn(s) + 2HCL (aq) -----------> ZnCl2(aq) + H2(g)

How many milliliters of 2.00 M HCl(aq) are required to react with 7.05 g of an ore containing 38.0% Zn(s) by mass?

Answer #1

Zinc reacts with hydrochloric acid according to the reaction equation

Zn(s) + 2HCL (aq) -----------> ZnCl2(aq) + H2(g)

First calculate the mass of Zn which is 38.0% Zn(s) by mass of ore.

Mass of Zn / total mass of Ore = % of Zn

Mass of Zn = 38.0/100* 7.05 g

= 2.679 g

Now calculate the moles of Zn:

2.679 g Zn x (1 mole Zn / 65.4 g Zn) = 0.041 moles Zn

Now calculate the moles of HCl:

0.041 moles Zn x (2 moles HCl / 1 mole Zn) = 0.082 moles HCl

moles solute = Molarity solute x L solution

Now calculate the volume of HCl in L as follows:

moles HCl = M HCl x L HCl

0.082 = (2.00)(L HCl)

L HCl = 0.082/ 2.00

= 0.041 L

- L = 1000 mL

**0.041 L = 41 ml**

Hence 41 milliliters of 2.00 M HCl(aq) are required to react with 7.05 g of an ore containing 38.0% Zn(s) by mass.

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