Question

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 °c is __________ l

Answer 1

Answer:

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

Answer 2

Answer:

13.6 L

Explanation:

Complete reaction: HCl + Mg → MgCl2 + H2

Balanced: 2HCl + Mg → MgCl2 + H2

Mole to mole comparison: 2 mole of HCl forms 1 mol of H2

If 2 moles of HCl are required to produce 1 mole of H2, then we need 2 times the amount of hydrogen gas given to get our answer of liters of HCl.

Hydrogen gas: 6.82 L

HCl : 6.82 x 2 = 13.6 L

ANSWER: 13.6 L