1. Calculate the relative formula masses (M_r):
Mg: 24.31 g/mol
HCl: 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
MgCl2: 24.31 g/mol (Mg) + 2
35.45 g/mol (Cl) = 95.21 g/mol
H2: 2
1.01 g/mol (H) = 2.02 g/mol
2. Calculate the moles of magnesium:
Moles of Mg = mass / M_r = 1 g / 24.31 g/mol = 0.0412 mol
3. Calculate the moles of hydrogen produced:
From the balanced equation, 1 mole of Mg produces 1 mole of H2. Therefore, 0.0412 mol of Mg will produce 0.0412 mol of H2.
4. Calculate the volume of hydrogen using the ideal gas law:
PV = nRT
Where:
P = pressure = 1 atm
V = volume (what we want to find)
n = moles of H2 = 0.0412 mol
R = ideal gas constant = 0.0821 L atm/mol K
T = temperature = 25°C = 298 K
V = nRT / P = (0.0412 mol
0.0821 L atm/mol K
298 K) / 1 atm = 1.01 L
Therefore, 1 g of magnesium reacting with excess hydrochloric acid will produce approximately 1.01 liters of hydrogen gas at 1 atmosphere and 25°C.