The volume of hcl gas required to react with excess magnesium metal to produce 6.82 l of hydrogen gas at 2.19 atm and 35.0 â°c is l

3.41LHCl

Explanation:

Hello! Let's solve this!

First we must raise the balanced equation

2HCl + Mg > H2 + MgCl2

From the equation we can see that the ratio of moles is 2: 1

2moles of HCl per 1 mol of H2

We know that the volume of H2 is 6.82L

So we follow the following relationship

6.82LH2 * (1mol H2 / 2molHCl) = 3.41LHCl

it is the most commonly accepted one