What is 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?

2 hcl(g) + mg(s) ? mgcl2(s) + h2(g)

Explanation:

2 HCl(g) + Mg(s) → MgCl₂(s) + H₂(g)

Let's calculate the quantity of mole of produced hydrogen with the Ideal Gases Law

P . V = n . R .T

2.19 atm . 6.82L = n . 0.082 . 308K

(2.19 atm . 6.82L) / (0.082 . 308K) = n

0.591 mol = n

1 mol of H₂ gas came from 2 mol of hydrochloric, so, 0.591 mol came from the double of mole

0.591 .2 = 1.182 mole of acid.

Molar mass of HCl = 36.45 g/m

1.182 mole are (36.45 g/m . 1.182g ) contained in 43.1 g

Density HCl = HCl mass / HCl volume

0,118 g/mL = 43.1 g / HCl volume

43.1 g / 0.118 g/mL = 365.3 mL (HCl volume)

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.