To determine the volume of hydrogen gas (H2) produced in the reaction between magnesium (Mg) and hydrochloric acid (HCl), we can start by looking at the balanced chemical equation:
Mg + 2 HCl → MgCl2 + H2
This equation tells us that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. We need to calculate how many moles of magnesium and hydrochloric acid we have, and then see which one is the limiting reactant.
Step 1: Calculate moles of Mg
The molar mass of magnesium (Mg) is approximately 24.31 g/mol. Using this, we can calculate the number of moles of magnesium:
Moles of Mg = Mass of Mg / Molar mass of Mg = 50.0 g / 24.31 g/mol ≈ 2.06 moles
Step 2: Calculate moles of HCl
The molar mass of hydrochloric acid (HCl) is approximately 36.46 g/mol. Now we can calculate the number of moles of HCl:
Moles of HCl = Mass of HCl / Molar mass of HCl = 75 g / 36.46 g/mol ≈ 2.06 moles
Step 3: Determine the limiting reactant
According to the balanced equation, 2 moles of HCl are required for every 1 mole of Mg. We calculated that we have approximately 2.06 moles of each:
– Mg: 2.06 moles
– HCl: 2.06 moles
To fully react 2.06 moles of Mg, we would need:
Required moles of HCl = 2 * moles of Mg = 2 * 2.06 = 4.12 moles
Since we only have 2.06 moles of HCl available, HCl is the limiting reactant.
Step 4: Calculate the amount of H2 produced
From our balanced equation, we know:
– 2 moles of HCl produce 1 mole of H2
Thus, 2.06 moles of HCl will produce:
Moles of H2 = (2.06 moles HCl) / 2 = 1.03 moles H2
Step 5: Convert moles of H2 to volume at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of hydrogen gas produced is:
Volume of H2 = moles of H2 * 22.4 L/mol = 1.03 moles * 22.4 L/mol ≈ 23.07 L
Conclusion
In conclusion, when 50.0 g of magnesium reacts with 75 g of hydrochloric acid, approximately 23.07 liters of hydrogen gas are produced at STP.