A mixture of .158 moles of C is reacted with .117 moles of O2 in a sealed 10L co

Question

A mixture of .158 moles of C is reacted with .117 moles of O2 in a sealed 10L container at 500K producing CO and CO2. The total pressure is .640 atm. What is the partial pressure of CO? The reaction is 3C + 2O2 ---> 2CO + CO2 Please don't forget to use the temperature and volume! Other answers to questions like this one didn't use it and their answers were wrong. Thank you!

Explanation / Answer

A mixture of 0.164 moles of C is reacted with 0.117 moles of O2 is a sealed, 10.0L vessel at 500K, producing a mixture of CO and CO2. The total pressure is 0.640atm. What is the partial pressure of CO? 3C(s)+2O2(g)---->2CO(g)+CO2(g) Expected ratio of no of moles of C to O2 is 3:2 but the ratio of the C available to the O2 available is 1.4:1. Since there is more O2 than required, C is the limiting reagent, implying that all of it will react. No of moles of CO produced= 2/3(no of moles of C) = 2/3(0.164) = 0.109 We now use the formula PV= nRT, where P is the partial pressure of CO, V is the volume of CO, n is the no of moles of CO, T is the temperature and R is the molar gas constant(8.31J/mol/K) P= nRT/V = (0.109)(8.31)(500)/(0.01) = 4.53x10^4N/m^2 1.01x10^5N/m^2= 1atm 4.53x10^4N/m^2= (4.53x10^4)/(1.01x10^5) = 0.449atm(partial pressure of CO)

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