The equation $4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O$ , represents the catalytic oxidation of ammonia. If $100 {c m}^{3}$ of ammonia is used, calculate the volume of oxygen required to oxidise the ammonia completely.

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Solution

$4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$
4 vol. : 5 vol. : 4vol. : 6 vol.
If 4 vol. of $N H_{3}$ requires 5 vol. of $O_{2}$
Then, 100 $c m^{3}$ of $N H_{3}$ require
= $\frac{5}{4} \times 100$
= 125 g of $O_{2}$

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The equation 4NH3+5O2⟶4NO+6H2O, represents the catalytic oxidation of ammonia. If 100cm3 of ammonia is used, calculate the volume of oxygen required to oxidise the ammonia completely.

4NH3+5O2→4NO+6H2O4 vol. : 5 vol. : 4vol. : 6 vol.If 4 vol. of NH3 requires 5 vol. of O2Then, 100cm3 of NH3require=54×100= 125 g of O2

The equation $4 N H_{3} + 5 O_{2} ⟶ 4 N O + 6 H_{2} O$ , represents the catalytic oxidation of ammonia. If $100 {c m}^{3}$ of ammonia is used, calculate the volume of oxygen required to oxidise the ammonia completely.

$4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$
4 vol. : 5 vol. : 4vol. : 6 vol.
If 4 vol. of $N H_{3}$ requires 5 vol. of $O_{2}$
Then, 100 $c m^{3}$ of $N H_{3}$ require
= $\frac{5}{4} \times 100$
= 125 g of $O_{2}$