How to do this chemistry problem with gases?

Chemistry Problem- Gases?

• A steel container contains 5.00 mol of graphite (pure carbon and 5.00 mol of O2). The mixture is ignited and all the graphite reacts. Combustion produces a mixture of CO gas and CO2 gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0 %. Calculate the mole fractions of CO2 and O2 in the final gaseous mixture

• Answer:

  5.00 mol gas + 17% of 5.00 = 5.85 mol of gas C(s) + O2 --> CO2 2C(s) + O2 --> 2CO production of CO2 does not increase the moles of gas (you get 1 mole of CO2 for each mole of O2 consumed) production of CO yield a net increase of 1 mole of gas (+2 mole CO for 1 mole of O2 consumed) +1.70 mol CO consumes 0.85 mole O2 At the end you have 1.70 mole CO and 4.15 mol of CO2+O2 mole fraction of CO = 0.29 (1.70 mol/5.85 mol) mole fraction of O2+CO2 = 0.71 (4.15 mol/5.85 mol) You will have O2 remaining because C is limiting You know that you used 1.70 mole of C to produce 1.70 mol of CO (using 0.85 mol of O2) so you have only 3.30 mol of C left to produce 3.30 mol CO2 (using 3.30 mol O2) Moles of O2 left = 5.00 - 0.85 - 3.30 = 0.85 mole fraction O2 = 0.85/5.85 = 0.145 mole fraction CO2 = 3.30/5.85 = 0.565 mole fraction CO = 0.290