Question

Consider the following reaction:

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

If 1.52 g of solid magnesium reacts with 100.0 mL of 3.00 M hydrochloric acid, what volume of hydrogen gas is produced at 23°C and 1.00 atm?

Answer 1

Approximately 1.52 liters of hydrogen gas is produced when 1.52 g of magnesium reacts with hydrochloric acid at 23°C and 1.00 atm.

Step-by-step explanation:

The student wants to know the volume of hydrogen gas produced when 1.52 g of solid magnesium reacts with hydrochloric acid (HCl) at a specified temperature and pressure.

Solution

First, calculate the number of moles of magnesium:

Molar mass of Mg = 24.305 g/mol

Moles of Mg = mass (g) / molar mass (g/mol) = 1.52 g / 24.305 g/mol ≈ 0.0625 mol

Using the stoichiometry of the reaction, Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g), each mole of Mg produces one mole of H₂. Therefore, 0.0625 mol of Mg will produce 0.0625 mol of H₂.

Using the ideal gas law, PV = nRT, and converting temperature to Kelvin (T=23°C+273=296K), we can solve for V (volume of H₂ at 1 atm and 296 K).

R (ideal gas constant) = 0.0821 L·atm/mol·K

V = nRT/P

V = (0.0625 mol) * (0.0821 L·atm/mol·K) * (296 K) / (1.00 atm)

V ≈ 1.52 L

Therefore, the volume of hydrogen gas produced is approximately 1.52 liters.