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

The volume of HCl gas required to react with excess magnesium metal to produce 6.82 L of hydrogen gas can be calculated using stoichiometry and the Ideal Gas Law. The volume of HCl gas required is approximately 6.82 L.

Step-by-step explanation:

To find the volume of HCl gas required, we need to use the stoichiometry of the reaction between magnesium and HCl. The balanced equation for this reaction is:

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

According to the equation, 1 mole of magnesium reacts with 2 moles of HCl to produce 1 mole of hydrogen gas. Given that 6.82 L of hydrogen gas is produced, we can use the stoichiometry to find the volume of HCl gas required.

1 mole of H2 = 22.4 L at STP (Standard Temperature and Pressure)

So, 6.82 L of H2 is equivalent to (6.82/22.4) moles of H2.

According to the balanced equation, 1 mole of H2 is produced from 2 moles of HCl.

Therefore, (6.82/22.4) moles of H2 would require ((6.82/22.4) x 2) moles of HCl.

Now, we can use the Ideal Gas Law to convert moles of HCl to volume:

Moles of HCl = Volume of HCl / (22.4 L/mol)

Volume of HCl = Moles of HCl x 22.4 L/mol

Replace Moles of HCl with ((6.82/22.4) x 2) moles, we get:

Volume of HCl = ((6.82/22.4) x 2) x 22.4 L/mol

Simplifying the equation, we find that the volume of HCl gas required is approximately 6.82 L.

Answer 2

13.6 L

Step-by-step 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