Could someone help me please

Question

Zinc reacts with hydrochloric acid according to the reaction equation shown. Zn(s) + 2 HCI(ag) - > ZnCl2(aq) + H2(g) How many milliliters of 6.50 M HCl(ag) are required to react with 9.35 g of an ore containing 25.0% Zn(s) by mass?

Answer 1

First, let's write a correctly balanced equation for the reaction that is taking place:

Zn(s) + 2HCl(aq) ==> ZnCl2(aq) + H2(g) ... balanced equation

Next, we will find the MOLES of Zn that are present:

Atomic mass Zn = 65.39 g / mole

9.35 g of ore x 25.0% Zn = 2.338 g of Zn

2.338 g Zn x 1 mol Zn / 65.39 g = 0.0357 moles Zn in the ore sample

Next, we will use the stoichiometry (mole ratio) of the balanced equation to find the MOLES of HCl needed:

0.0357 mols Zn x 2 mols HCl / 1 mol Zn = 0.0714 mols HCl needed

Finally, using the given concentration of HCl (6.5 M), we will convert mols HCl to mls HCl:

6.5 M HCl mean 6.5 moles HCl / L

0.0714 mols HCl x 1 L / 6.5 mols = 0.01098 L

0.01098 L x 1000 mls / L = 10.98 mls = 11.0 mls (to 3 sig.figs.) = Answer