Question

Calculate the mass of Nitric oxide produced by the reaction of 2 mole of

ammonia with 2 moles of oxygen​

Answer 1

Explanation:

2NO+O

2

⟶2NO

2

Number of moles of NO in 20.3g=

30

20.3g

=0.67 moles

Number of moles of O

2

in 5.6g=

32

5.6

=0.175 moles

As 1 mole of O

2

is reacting with 2 moles of NO

0.175 moles of O

2

is reacting with 2×0.175 moles of NO=0.35 moles of NO .

Thus O

2

is limiting agent.

Moles 2NO+O

2

⟶2NO

2

at t=0 0.67 0.175 -

After reaction 0.67−0.35 - 0.35

Thus moles of NO

2

produce= 0.35 as 1 mole of O

2

give 2 moles of NO

2

.

Mass of NO

2

produced= n × Molar mass

=0.35×46

⇒m

NO

2

=16.1g

Thus 16.1g of NO

2

is produced.

Answer 2

Answer:

48 g of nitric oxide(NO).

Explanation:

The balanced chemical reaction is:

4NH3 + 5O2   ↔ 4NO + 6H2O

As we know that the molar masses of NH3(ammonia) and O2(oxygen) are 17 g/mol and 32 g/mol respectively.

5 moles (160 g) of O2 reacts with 4 moles (68 g) of NH3

2 moles of oxygen = 64 g of oxygen will react with $\frac{68\times64}{160}$ = 27.2 g of ammonia = 1.6 moles of ammonia.

​ Hence, oxygen is the limiting regent(as the total 2 moles of oxygen gets  consumed).

And the molar mass of NO is 30 g/mol.

5 moles (160 g) of oxygen will produce 4 moles (120 g) of NO.

64 grams of oxygen will produce  $\frac{120\times64}{160}$ = 48 g of nitric oxide(NO).

#SPJ2