Question

Hydrogen gas can be produced in the laboratory through the reaction of hydrochloric acid with magnesium metal. 2HCl(aq) + Mg(s) → H₂(g) + MgCl₂(aq). When 10.6 g of Mg reacts, what volume, in liters, of H₂ gas is produced at 24°C and 847 mmHg?

Answer 1

Calculate the moles of magnesium by dividing the mass by its molar mass. Convert the temperature to Kelvin and the pressure to atmospheres. Then apply the ideal gas law to find the volume of hydrogen gas produced.

Step-by-step explanation:

The student's question involves the production of hydrogen gas through the reaction of hydrochloric acid with magnesium metal. To calculate the volume of hydrogen gas produced when 10.6 g of Mg reacts at 24°C and 847 mmHg, we'll use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

First, we'll convert the given mass of Mg to moles using its molar mass, which is 24.305 g/mol for magnesium. Then we'll use the ideal gas law to find the volume. The pressure must be converted to atmospheres (using the conversion 1 atm = 760 mmHg), and the temperature must be converted to Kelvin (using the conversion °C + 273.15 = K).

Step 1: Calculate moles of Mg.

• Moles of Mg = Mass of Mg / Molar mass of Mg
  
• Moles of Mg = 10.6 g / 24.305 g/mol

Step 2: Convert temperature to Kelvin and pressure to atmospheres.

• T(K) = 24°C + 273.15 = 297.15 K
  
• P(atm) = 847 mmHg / 760 mmHg/atm

Step 3: Apply the ideal gas law to find the volume of H2.

• V = nRT / P

Having calculated the moles of Mg and converted the temperature and pressure to the correct units, we can calculate the volume of hydrogen gas produced. Remember that according to the stoichiometry of the given reaction, 1 mole of magnesium produces 1 mole of hydrogen gas.