Question

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)

Answer 1

Answer:

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)