Complete Question:
What volume of hydrogen will be produced at STP by the reaction 67.3 g of magnesium with excess water according to the following reaction?
Mg + 2H₂O --> Mg(OH)₂ + H₂
Answer:
62 L
Solution:
Step 1: Calculate Moles of Mg as;
Moles = Mass / M.Mass
Moles = 67.3 g / 24.30 g/mol
Moles = 2.76 moles of Mg
Step 2: Calculate Moles of H₂ as;
According to balance chemical equation,
1 mole of Mg produced = 1 mole of H₂
So,
2.76 moles of Mg will produce = X moles of H₂
Solving for X,
X = 2.76 mol × 1 mol / 1 mol
X = 2.76 mol
Step 3: Calculating volume of H₂,
1 mole of ideal H₂ occupies = 22.4 L Volume at STP
So,
�� 2.76 moles of H₂ will occupy = X L of H₂ at STP
Solving for X,
X = 2.76 mol × 22.4 L / 1 mol
X = 61.82 L ≈ 62 L
