# Question 6dead

Mar 1, 2015

You'd need $\text{940 g}$ of zinc metal to react completely with that particular $\text{HCl}$ solution.

The starting point will be your balanced chemical equation

$Z {n}_{\left(s\right)} + 2 H C {l}_{\left(a q\right)} \to Z n C {l}_{2 \left(a q\right)} + {H}_{2 \left(g\right)}$

Now look at the $\text{1:2}$ mole ratio that exists between zinc and hydrochloric acid; every mole of zinc needs 2 moles of hydrochloric acid for the reaction to take place.

This means that you can determine the number of moles of zinc that react by calculating the number of moles of hydrochloric acid that took part in the reaction.

You can do this by using the molarity of the $\text{HCl}$ solution

C = n/V => n_("HCl") = C * V = "3.5 M" * "8.2 L" = "28.7 moles" $\text{HCl}$

Using the mole ratio will get you

$\text{28.7 moles HCl" * "1 mole zinc"/"2 moles HCl" = "14.35 moles zinc}$

Calculate the mass of zinc by using its molar mass

$\text{14.35 moles" * "65.4 g"/"1 mole" = "938.5 g zinc}$

Rounded to two sig figs, the answer will be

m_("zinc") = "940 g"#