Answer :
To determine the volume of [tex]\text{H}_2[/tex] gas required to convert [tex]2.8 \text{ g}[/tex] of [tex]\text{N}_2[/tex] into [tex]\text{NH}_3[/tex], we'll follow these steps:
Find the number of moles of [tex]\text{N}_2[/tex]:
The molar mass of [tex]\text{N}_2[/tex] is approximately [tex]28 \text{ g/mol}[/tex]. Therefore, the number of moles of [tex]\text{N}_2[/tex] is given by:
[tex]\text{moles of } \text{N}_2 = \frac{2.8 \text{ g}}{28 \text{ g/mol}} = 0.1 \text{ mol}[/tex]
Use the balanced chemical equation:
The balanced chemical equation for the formation of ammonia [tex](\text{NH}_3)[/tex] is:
[tex]\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3[/tex]
From the equation, [tex]1 \text{ mole of } \text{N}_2[/tex] reacts with [tex]3 \text{ moles of } \text{H}_2[/tex].
Calculate the moles of [tex]\text{H}_2[/tex] required:
If [tex]0.1 \text{ mol of } \text{N}_2[/tex] is used, then:
- [tex]0.1 \text{ mol of } \text{N}_2[/tex] requires [tex]0.1 \times 3 = 0.3 \text{ mol of } \text{H}_2[/tex].
Convert moles of [tex]\text{H}_2[/tex] to volume at NTP:
At Normal Temperature and Pressure (NTP), [tex]1 \text{ mole of gas occupies 22.4 \text{ liters}}[/tex]. Therefore, the volume [tex]V[/tex] of [tex]\text{H}_2[/tex] is:
[tex]V = 0.3 \text{ mol} \times 22.4 \text{ L/mol} = 6.72 \text{ L}[/tex]
Since [tex]6.72 \text{ L} = 6720 \text{ mL}[/tex], the volume of [tex]\text{H}_2[/tex] required is [tex]6.72 \text{ L}[/tex], which corresponds to option 3).
Therefore, the correct choice is 6.72 lit (option 3).
