How many moles of oxygen are needed to react with 6.78 grams of ammonia?

4NH3 + 5O2-> 4NO + 6 H2O

Answer: 0.499mol

Explanation:

4NH3 + 5O2 - > 4NO + 6H2O

Molar Mass of NH3 = 14 + 3 = 17g/mol

Mass conc. Of NH3 = 6.78g

n = Mass conc. / molar Mass

n = 6.78 / 17 = 0.399mol

From the equation,

4moles of NH3 required 5moles of O2.

Therefore, 0.399mol of NH3 will require Xmole of O2. i. e

Xmol of O2 = (0.399 x 5) / 4 = 0.499mol