Answer :
To solve this problem, we'll use the Ideal Gas Law and stoichiometry. Let's break it down into two parts:
Part A: Determine moles of [tex]O_2[/tex]
Convert the temperature from Celsius to Kelvin:
[tex]T(K) = 75 + 273.15 = 348.15 \, \text{K}[/tex]
Use the Ideal Gas Law [tex]PV = nRT[/tex] to find the total moles of gas [tex]n[/tex]:
- [tex]P = 99.3 \, \text{kPa} = 99.3 \, \text{kPa} = 99,300 \, \text{Pa}[/tex]
- [tex]V = 525 \, \text{cm}^3 = 0.000525 \, \text{m}^3[/tex]
- [tex]R = 8.314 \, \text{J/(mol·K)}[/tex]
- [tex]T = 348.15 \, \text{K}[/tex]
Rearrange the Ideal Gas Law to solve for [tex]n[/tex]:
[tex]n = \frac{PV}{RT} = \frac{99,300 \, \text{Pa} \times 0.000525 \, \text{m}^3}{8.314 \, \text{J/(mol·K)} \times 348.15 \, \text{K}} \approx 0.0184 \, \text{mol}[/tex]
Calculate the moles of [tex]O_2[/tex]:
Since the mole fraction of [tex]O_2[/tex] in air is 0.209, the moles of [tex]O_2[/tex] is:
[tex]n_{O_2} = 0.209 \times 0.0184 \, \text{mol} \approx 0.00385 \, \text{mol}[/tex]
Part B: Calculate the mass of [tex]C_8H_{18}[/tex] that can be combusted
Balanced chemical equation for the combustion of octane ([tex]C_8H_{18}[/tex]):
[tex]2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O[/tex]
This tells us that 2 moles of [tex]C_8H_{18}[/tex] react with 25 moles of [tex]O_2[/tex].
Determine moles of [tex]C_8H_{18}[/tex] that can be combusted:
Using stoichiometry from the balanced equation:
[tex]n_{C_8H_{18}} = \frac{2}{25} \times n_{O_2}[/tex]
[tex]n_{C_8H_{18}} = \frac{2}{25} \times 0.00385 \, \text{mol} \approx 0.000308 \, \text{mol}[/tex]
Calculate the mass of [tex]C_8H_{18}[/tex]:
The molar mass of [tex]C_8H_{18}[/tex] is [tex](8 \times 12.01) + (18 \times 1.008) = 114.22 \, \text{g/mol}[/tex].
[tex]m = n \times \text{molar mass} = 0.000308 \, \text{mol} \times 114.22 \, \text{g/mol} \approx 0.0352 \, \text{g}[/tex]
Therefore, about 0.0352 grams of [tex]C_8H_{18}[/tex] could be combusted by the given quantity of [tex]O_2[/tex], assuming complete combustion.