The moles of O 2 required for reacting with 6.8 g ammonia. ( ldots . . NH 3+ ldots . O 2 arrow ldots . NO + ldots . H 2 O ) is:

Question

The moles of O 2 required for reacting with 6.8 g ammonia. ( ldots . . NH 3+ ldots . O 2 arrow ldots . NO + ldots . H 2 O ) is: (A) 5 (B) 2.5 (C) 1 (D) 0.5

Tardigrade Admin · Accepted Answer

The balanced equation is: 4 NH 3+5 O 2 arrow 4 NO +6 H 2 O ...(1) 4 × 17 g of NH 3 ( eq -1) Required O 2=5 mol 6.8 of NH 3 require O 2=(5 × 6.8/4 × 17) =0.5 mol

Q. The moles of $O_{2}$ required for reacting with $6.8 g$ ammonia. $(\dots .. N H_{3} + \dots . O_{2} \to \dots . NO + \dots . H_{2} O)$ is:

5

2.5

1

0.5

The balanced equation is:

$4 N H_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O$ ...(1)

$4 \times 17 g$ of $N H_{3}$ ( eq -1)

Required $O_{2} = 5 m o l$

$6.8$ of $N H_{3}$ require $O_{2} = \frac{5 \times 6.8}{4 \times 17}$

$= 0.5 m o l$

Q. The moles of O2​ required for reacting with 6.8g ammonia. (…..NH3​+….O2​→….NO+….H2​O) is:

5

2.5

1

0.5

The balanced equation is: 4NH3​+5O2​→4NO+6H2​O ...(1) 4×17g of NH3​ ( eq -1) Required O2​=5mol 6.8 of NH3​ require O2​=4×175×6.8​ =0.5mol

Q. The moles of $O_{2}$ required for reacting with $6.8 g$ ammonia. $(\dots .. N H_{3} + \dots . O_{2} \to \dots . NO + \dots . H_{2} O)$ is:

The balanced equation is:

$4 N H_{3} + 5 O_{2} \to 4 NO + 6 H_{2} O$ ...(1)

$4 \times 17 g$ of $N H_{3}$ ( eq -1)

Required $O_{2} = 5 m o l$

$6.8$ of $N H_{3}$ require $O_{2} = \frac{5 \times 6.8}{4 \times 17}$

$= 0.5 m o l$