What volumeof hydrogen at stp by the reaction of 67.3

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