Question

The number moles of $O_2$ required for reacting with $6.8$ g of ammonia is:

$N{H}_3+O_2\to NO+H_{2}O$

• A. $5$
• B. $2.5$
• C. $1$
• D. $0.5$

Answer 1

The correct option is

C

$0.5$

The balanced reaction is as follows:

NH3+54O2→NO+32H2O\displaystyle NH_{3}+\frac{5}{4}O_{2}\rightarrow NO+\frac{3}{2}H_{2

4NH3+5O2→4NO+6H2O\displaystyle 4NH_{3}+5O_{2}\rightarrow 4NO+6H_{2}O4NH3​+5O2​→4NO+6H2​O$....(i)$

5 moles of oxygen are required to react with a 4 mole of ammonia

From the equation, 4×17g of $N{H}_3$ required $O_2$ =5 mol

6⋅8g of $N{H}_3$ require $O_2$

$\frac{5\times 6.8}{4\times 17}=0.5mol$

x=5×6.84×17=0.5x=\dfrac{5 \times 6.8}{4 \times 17}=0.5

Hence, the number of moles of O2O_2O2​ required is 0.50.50.5 mol.

Hence, the correct option is CC C