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15240672380's Question
Chemistry
Posted 3 months ago

A steel cylinder contains 5.00 mole of graphite (pure carbon) and 5.00 moles of O2\mathrm{O}_{2}. The mixture is ignited and all the graphite reacts. Combustion produces a mixture of COC O gas and CO2\mathrm{CO}_{2} gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0%17.0 \%. Calculate the mole fractions of CO,CO2\mathrm{CO}, \mathrm{CO}_{2}, and O2\mathrm{O}_{2} in the final gaseous mixture.
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Answer from Sia
Posted 3 months ago
Solution
1
Write the balanced chemical equations: The combustion of carbon (graphite) can produce either carbon monoxide (COCO) or carbon dioxide (CO2CO_2). The two possible reactions are: C+O2CO2C + O_2 \rightarrow CO_2 and 2C+O22CO2C + O_2 \rightarrow 2CO
2
Determine the limiting reactant: Since we have equal moles of CC and O2O_2, and the reaction to form CO2CO_2 requires a 1:1 mole ratio, while the reaction to form COCO requires a 2:1 mole ratio of CC to O2O_2, O2O_2 is the limiting reactant for both reactions
3
Calculate the moles of gases after reaction: If all O2O_2 is used to form CO2CO_2, we would have 5 moles of CO2CO_2. If all O2O_2 is used to form COCO, we would have 10 moles of COCO. Since the pressure increased by 17%, we know that the total moles of gas increased. This means that some O2O_2 must have been used to form COCO. Let xx be the moles of O2O_2 that reacted to form CO2CO_2, then 5x5 - x is the moles of O2O_2 that reacted to form COCO. The total moles of gas after reaction is x+2(5x)+(5x)=10+xx + 2(5 - x) + (5 - x) = 10 + x
4
Relate the change in moles to change in pressure: Assuming ideal gas behavior and constant temperature and volume, the pressure is proportional to the number of moles of gas. Let P0P_0 be the original pressure, then the final pressure is P0×1.17P_0 \times 1.17. The initial moles of gas is 10 (5 moles of CC and 5 moles of O2O_2), so 10×1.17=10+x10 \times 1.17 = 10 + x. Solving for xx gives x=1.7x = 1.7 moles
5
Calculate the mole fractions: The mole fraction of a component is the number of moles of that component divided by the total number of moles. The mole fraction of CO2CO_2 is x10+x\frac{x}{10 + x}, the mole fraction of COCO is 2(5x)10+x\frac{2(5 - x)}{10 + x}, and the mole fraction of O2O_2 is 5x10+x\frac{5 - x}{10 + x}. Plugging in x=1.7x = 1.7 gives the mole fractions
Answer
Mole fraction of CO2CO_2: 1.711.70.145\frac{1.7}{11.7} \approx 0.145
 Mole fraction of COCO: 2(51.7)11.70.568\frac{2(5 - 1.7)}{11.7} \approx 0.568
Mole fraction of O2O_2: 51.711.70.282\frac{5 - 1.7}{11.7} \approx 0.282
Key Concept
The mole fraction of a component in a mixture is the ratio of the number of moles of that component to the total number of moles of all components in the mixture.
Explanation
The mole fractions were calculated based on the stoichiometry of the combustion reactions and the given increase in pressure, which indicates the total moles of gas after the reaction. By assuming ideal gas behavior, the change in pressure directly relates to the change in the number of moles of gas.

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