Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g) How many milliliters of 5.50 M HCl(aq) are required to react with 7.35 g Zn(s)?

Answer 1

Zn(s) + 2HCl(aq) ⟶ ZnCl₂(aq) + H₂(g) ... BALANCED EQUATION

From the balanced equation we see that it requires TWO moles of HCl for each ONE mole of Zn. So, we will find the moles of Zn, then the moles of HCl and finally the milliliters of HCl.

moles Zn present = 7.35 g Zn x 1 mol Zn/65.4 g = 0.1134 mole Zn

moles HCl needed = 0.1134 mol Zn x 2 mol HCl/mol Zn = 0.2248 mol HCl needed

mls HCl required: (x L)(5.50 mol/L) = 0.2248 mol and x = 0.0409 L x 1000 ml/L = 40.9 mls (3 sig. figs.)