Answer to Question #144568 in Chemistry for tala

Question

Consider the following equation: Mg(s)+2H₂O(l)→Mg(OH)₂(s)+H₂(g). Find the volume of H₂ produced at STP by the reaction of 76.4 g of magnesium. *

Answer 1

At Standard Temperature and Pressure (STP), 1 mole of any gas will occupy a volume of 22.4 L (V_m).

Solution:

The balanced chemical equation:

Mg(s) + 2H₂O(l) → Mg(OH)₂(s) + H₂(g)

According to the equation: n(Mg) = n(H₂)

Moles of Mg = n(Mg) = Mass Mg / Molar mass of Mg

The molar mass of Mg is 24.305 g mol^-1

Thus,

n(Mg) = m(Mg) / M(Mg) = 76.4 g / 24.305 g mol^-1 = 3.1434 mol

n(Mg) = 3.1434 mol

Hence,

n(H₂) = n(Mg) = 3.1434 mol

Moles of H₂ = n(H₂) = Volume of H₂ / V_m

Volume of H₂ = n(H₂) × V_m = 3.1434 mol × 22.4 L mol^-1 = 70.412 L = 70.4 L

V(H₂) = 70.4 L

Answer: 70.4 L of H₂