Respuesta :
Explanation:
Chemical reaction equation for the given reaction is as follows.
[tex]2KNO_{3}(s) + \frac{1}{8}S_{8}(s) + 3C(s) \rightarrow K_{2}S(s) + N_{2}(g) 3CO_{2}(g)[/tex]
Therefore, we will calculate the number of moles of [tex]KNO_{3}[/tex] as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{4.05 g}{101.1 g/mol}[/tex]
= 0.04 mol
Number of moles of nitrogen gas formed is calculated as follows.
No. of moles = [tex]\frac{1}{2} \times \text{no. of moles of KNO_{3}}[/tex]
= [tex]\frac{1}{2} \times 0.04 mol[/tex]
= 0.02 mol
Mass of [tex]N_{2}[/tex] gas formed will be calculated as follows.
No. of moles × Molar mass of [tex]N_{2}[/tex]
= [tex]0.02 mol \times 28.0 g/mol[/tex]
= 0.56 g
Now, the number of moles of [tex]CO_{2}[/tex] formed is as follows.
= [tex]\frac{3}{2} \times 0.04[/tex]
= 0.06 mol
Hence, mass of [tex]CO_{2}(g)[/tex] formed will be as follows.
[tex]0.04 mol \times 44 g/mol[/tex]
= 1.76 g
Volume of [tex]N_{2}(g)[/tex] is calculated as follows.
Volume = [tex]\frac{mass}{density}[/tex]
= [tex]\frac{0.56 g}{1.165 g/L}[/tex]
= 0.48 L
And, volume of [tex]CO_{2}[/tex] is calculated as follows.
Volume = [tex]\frac{mass}{density}[/tex]
= [tex]\frac{1.76 g}{1.830 g/L}[/tex]
= 0.96 L
Let us assume that the volume of solids are negligible. Therefore, total volume will be as follows.
[tex]\Delta V[/tex] = (0.48 L + 0.96 L)
= 1.44 L
Relation between work, pressure and volume is as follows.
w = -[tex]P_{ext} \times \Delta V[/tex]
= -[tex]1.00 atm \times 1.44 L[/tex]
= -1.44 atm L
As 1 tm L = 101.3 J. So, convert 1.44 atm L into joules as follows.
[tex]1.44 \times 101.3 J[/tex]
= 145.87 J
Thus, we can conclude that the given gases will do 145.87 J of work.
