Question

The moles of O2 required for reacting with 6.8 g of ammonia.......
(...NH3 + …O2 ----->…NO +…H2O)

(A). 0.5
(B). 2.5
(C). 1
(D). 5



Give answer with sol. plzz..

Answer 1

Answer : The correct option is A.

Solution : Given,

Mass of ammonia = 6.8 g

Molar mass of ammonia = 17.031 g/mole

First we have to calculate the moles of ammonia.

$\text{ Moles of }NH_3=\frac{\text{ Given mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{6.8g}{17.031g/mole}=0.399moles$

Now we have to calculate the moles of $O_2$ required.

The balanced chemical reaction is,

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

From the reaction, we conclude that

4 moles of $NH_3$ react with 5 moles of $O_2$

0.399 moles of $NH_3$ react with $\frac{5moles}{4moles}\times 0.399moles=0.498moles\approx 0.5moles$ of $O_2$

Therefore, the moles of $O_2$ required is 0.5 moles.

Answer 2

Answer: A) 0.5

Explanation:

Mass of ammonia = 6.8 g

Molar mass of ammonia = 17 g/mole

$\text{ Moles of }NH_3=\frac{\text{ Given mass of }NH_3}{\text{ Molar mass of }NH_3}=\frac{6.8g}{17.031g/mole}=0.4moles$

The balanced chemical reaction is,

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

According to stoichiometry:

4 moles of ammonia react with 5 moles of oxygen

0.4 moles of ammonia react with =$\frac{5}{4}\times 0.4=0.5$moles of oxygen

Thus correct answer is 0.5 moles.