Question

Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor. write balanced chemical equation, how many moles of ammonia will react with 6.73g oxygen?

Answer 1

Answer :

A balanced chemical equation for reaction between ammonia gas (NH3), Oxygen (O2), to yield Nitrogen monoxide (NO) and water vapor (H2O).

4 NH₃ (g) + 5 O₂ (g) ----------> 4 NO (g) + 6 H₂O (g).

(if you have doubt in getting balanced equation please let me know)

From above stoichiometrically balanced chemical equation it's clear that,

4 moles of NH3 ≡ 5 moles of O2 ≡ 4 moles of NO ≡ 6 moles of H2O.

Molar mass of O2 = 32 g/mole.

I.e.      32 g O2 ≡ 1 mole O2

Then, 6.73 g O2 ≡ say ‘A’ moles of O2.

On cross multiplication,

A x 32 = 1 x 6.37

A = 6.37 / 32

A = 0.199 moles.

Now from above mole relation we write,

If,         5 moles of O2 ≡ 4 moles of NH3

Then,    0.199 moles of O2 ≡ say ‘B’ moles pf NH3.

On cross multiplication,

B x 5 = 0.199 x 4

B = 0.199 x 4 / 5

B = 0.159 moles of NH3.

Answer: 6.73 g i.e. 0.199 mole of O2 will react completely with 0.159 mole of NH3.

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