Respuesta :
Answer : The correct option is, (5) [tex]N_2O,N_2O_4[/tex]
Explanation :
For 1st experiment :
First we have to calculate the moles of [tex]N_2[/tex].
Molar mass of [tex]N_2[/tex] = 28 g/mole
[tex]\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{3.62g}{28g/mole}=0.129moles[/tex]
Now we have to calculate the moles of [tex]O_2[/tex].
Molar mass of [tex]O_2[/tex] = 32 g/mole
[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{2.07g}{32g/mole}=0.0647moles[/tex]
Now we have to calculate the ratio of [tex]N_2\text{ and }O_2[/tex].
[tex]\frac{N_2}{O_2}=\frac{0.129}{0.0647}=1.99:1\approx 2:1[/tex]
Thus, the molecular formula of the nitrogen oxide will be, [tex]N_2O[/tex].
For 2nd experiment :
First we have to calculate the moles of [tex]N_2[/tex].
Molar mass of [tex]N_2[/tex] = 28 g/mole
[tex]\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{1.82g}{28g/mole}=0.065moles[/tex]
Now we have to calculate the moles of [tex]O_2[/tex].
Molar mass of [tex]O_2[/tex] = 32 g/mole
[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{4.13g}{32g/mole}=0.129moles[/tex]
Now we have to calculate the ratio of [tex]N_2\text{ and }O_2[/tex].
[tex]\frac{N_2}{O_2}=\frac{0.065}{0.129}=0.50:1[/tex]
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, [tex]NO_2\text { or }N_2O_4[/tex].
Hence, the correct option is, (5) [tex]N_2O,N_2O_4[/tex]
