Answer to Question #331770 in Chemistry for Hama

Question

How many liters of H, gas at STP can be produced

by the reaction of 4.60 g of Na and excess water

Answer 1

Solution:

The molar mass of Na is 23 g/mol

Therefore,

Moles of Na = (4.60 g Na) × (1 mol Na / 23 g Na) = 0.20 mol Na

Balanced chemical equation:

2Na + 2H₂O → 2NaOH + H₂

According to stoichiometry:

2 mol of Na produce 1 mol of H₂

Thus, 0.20 mol of Na produce:

(0.20 mol Na) × (1 mol H₂ / 2 mol Na) = 0.1 mol H₂

At STP, one mole of any gas occupies a volume of 22.4 L.

Thus, 0.1 mol of H₂ occupies:

(0.1 mol H₂) × (22.4 L / 1 mol) = 2.24 L H₂

Volume of H₂ = 2.24 L

Answer: 2.24 liters of H₂