Answer to Question #181132 in Chemistry for Taha

Question

calculate the number of moles of sodium oxide that should be produced when 26.50 grams of sodium metal reacts with oxygen gas. show work!

Answer 1

Solution:

Balanced chemical equation:

4Na(s) + O₂(g) → 2Na₂O(s)

According to the equation above: n(Na)/2 = n(Na₂O)

Moles of Na = Mass of Na / Molar mass of Na

The molar mass of Na is 23 g mol^-1.

Hence,

Moles of Na = 26.50 g / 23 g mol^-1= 1.152 mol

n(Na) = 1.152 mol

n(Na₂O) = n(Na)/2 = 1.152 / 2 = 0.576 mol

Moles of Na₂O = 0.576 mol

OR (short form solution):

(26.50 g Na) × (1 mol Na / 23 g Na) × (2 mol Na₂O / 4 mol Na) = 0.576 mol Na₂O

Answer: 0.576 moles of sodium oxide (Na₂O) should be produced.