Answer:
Part A
The volume of the gaseous product is [tex]V = 787L[/tex]
Part B
The volume of the the engine’s gaseous exhaust is [tex]V_e = 2178 \ L[/tex]
Explanation:
Part A
From the question we are told that
The temperature is [tex]T = 350^oC = 350 +273 =623K[/tex]
The pressure is [tex]P = 735 \ torr = \frac{735}{760} = 0.967\ atm[/tex]
The of [tex]C_8 H_{18} = 100.0g[/tex]
The chemical equation for this combustion is
[tex]2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}[/tex]
The number of moles of [tex]C_8 H_{18}[/tex] that reacted is mathematically represented as
[tex]n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }[/tex]
The molar mass of [tex]C_8 H_{18}[/tex] is constant value which is
[tex]M = 114.23 \ g/mole[/tex]
So [tex]n = \frac{100 }{114.23} }[/tex]
[tex]n = 0.8754 \ moles[/tex]
The gaseous product in the reaction is [tex]CO_2_{(g)}[/tex] and water vapour
Now from the reaction
2 moles of [tex]C_8 H_{18}[/tex] will react with 25 moles of [tex]O_2[/tex] to give (16 + 18) moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So
1 mole of [tex]C_8 H_{18}[/tex] will react with 12.5 moles of [tex]O_2[/tex] to give 17 moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
This implies that
0.8754 moles of [tex]C_8 H_{18}[/tex] will react with (12.5 * 0.8754 ) moles of [tex]O_2[/tex] to give (17 * 0.8754) of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So the no of moles of gaseous product is
[tex]N_g = 17 * 0.8754[/tex]
[tex]N_g = 14.88 \ moles[/tex]
From the ideal gas law
[tex]PV = N_gRT[/tex]
making V the subject
[tex]V = \frac{N_gRT}{P}[/tex]
Where R is the gas constant with a value [tex]R = 0.08206 \ L\cdot atm /K \cdot mole[/tex]
Substituting values
[tex]V = \frac{14.88* 0.08206 *623}{0.967}[/tex]
[tex]V = 787L[/tex]
Part B
From the reaction the number of moles of oxygen that reacted is
[tex]N_o = 0.8754 * 12.5[/tex]
[tex]N_o = 10.94 \ moles[/tex]
The volume is
[tex]V_o = \frac{10.94 * 0.08206 *623}{0.967}[/tex]
[tex]V_o = 579 \ L[/tex]
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
[tex]V_e = V_o * \frac{0.79}{0.21}[/tex]
Substituting values
[tex]V_e = 579 * \frac{0.79}{0.21}[/tex]
[tex]V_e = 2178 \ L[/tex]
