Question

Nitrogen and oxygen can react to form various compounds. Two experiments showed that one compound is formed when 3.62 g of nitrogen and 2.07 g of oxygen react completely, while another compound is formed when 1.82 g of nitrogen reacts completely with 4.13 g of oxygen. Which of the following are most likely the molecular formulas for the nitrogen oxides obtained in these experiments? (1) NO, N2O (2) NO, NO2 (3) N2O, N2O5 (4) NO, N2O4 (5) N2O, N2O4

Answer 1

Answer : The correct option is, (5)
`N_2O,N_2O_4`

Explanation :

For 1st experiment :

First we have to calculate the moles of
`N_2`.

Molar mass of
`N_2` = 28 g/mole

`\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=(3.62g)/(28g/mole)=0.129moles`

Now we have to calculate the moles of
`O_2`.

Molar mass of
`O_2` = 32 g/mole

`\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(2.07g)/(32g/mole)=0.0647moles`

Now we have to calculate the ratio of
`N_2\text{ and }O_2`.

`(N_2)/(O_2)=(0.129)/(0.0647)=1.99:1\approx 2:1`

Thus, the molecular formula of the nitrogen oxide will be,
`N_2O`.

For 2nd experiment :

First we have to calculate the moles of
`N_2`.

Molar mass of
`N_2` = 28 g/mole

`\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=(1.82g)/(28g/mole)=0.065moles`

Now we have to calculate the moles of
`O_2`.

Molar mass of
`O_2` = 32 g/mole

`\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(4.13g)/(32g/mole)=0.129moles`

Now we have to calculate the ratio of
`N_2\text{ and }O_2`.

`(N_2)/(O_2)=(0.065)/(0.129)=0.50:1`

To make a whole number, we are multiplying the ratio by 2, we get the ratio 1 : 2.

Thus, the molecular formula of the nitrogen oxide will be,
`NO_2\text { or }N_2O_4`.

Hence, the correct option is, (5)
`N_2O,N_2O_4`