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 (DON'T FORGET TO USE THE TEMPERATURE AND VOLUME OF THE CONTAINER! Other answers for a similar question did not include these factors and the answer was wrong). Thank you.

Explanation / Answer

PCO = mole fraction CO x Ptotal.. you know Ptotal.. you need mole fraction CO.. so you need to do a little stoichiometry to determine limiting reagent and amount of CO, CO2, and O2 when the reaction is complete..ok? ********** we have this reaction 3 C + 2 O2 ---> 2 CO + 1 CO2.. right? and if we convert mass C to moles CO 0.158g C x (1 mol C / 12.011g C) x (2 mol CO / 3 mol C) = 0.008770 and if we convert mass O to moles CO 0.117g O2 x (1 mol O2 / 32.00g O) x (2 mol CO / 2 mol O2) = 0.005513 since the O2 gave less CO, the O2 was limiting the reaction in which case.. 0.005513 moles CO was produced and moles of CO2 produced.... 0.117g O2 x (1 mol O2 / 32.00g O) x (1 mol CO2 / 2 mol O2) = 0.002766 assuming the remaining C was solid total moles of gas produced = 0.005513 + 0.002766 = 0.008297 and mole fraction CO in the gas = 0.005513 / 0.008297 = 2/3 ***** now we know.. PCO = 2/3 x Ptotal = 2/3 x 0.640 atm = 0.427 atm

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