in the combustion of 0.200 moles of pentane gas at 100 degrees celsius and 675 mm Hg in an excess of oxygen gas, how many liters of carbon dioxide gas can be generated?
2006-10-24 07:10:26 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Oxygen is in excess so pentane (C5H12) reacts fully. According to the balanced chem. equation:
C5H12 + 8O2 --> 5CO2 + 6H2O
0.2 mol of C5H12 produce 5*0.2 = 1 mol of CO2.
We can find the volume of CO2 from the state equation of the ideal gases:
p*V = n*R*T where p = 675/760 = 0.888 atm, n = 1 mol, R = 0.082 L*atm/mol*K and T = 100 + 273 = 373 K. Solving for the volume V we get:
V = n*R*T/p, V = 1*0.082*373/0.888, V = 34.44 L
2006-10-24 08:15:33 · answer #1 · answered by Dimos F 4 · 0⤊ 0⤋
Each mole of pentane can produce 5 moles of co2 because each molecule contains 5 carbons. Then use the ideal gas law, PV = nRT. Make sure you get your units right for the R that you use.
2006-10-24 14:20:20 · answer #2 · answered by Geoffrey S 3 · 0⤊ 0⤋
Set it up by writing the balanced equation:
C5H12 + 8O2 --> 5CO2 + 6H2O
If 0.2 moles of C5H12 are used, 5 times as much CO2 is formed according to the balanced equation. Use the ideal gas law (PV = nRT) and plug in the numbers.
2006-10-24 14:20:11 · answer #3 · answered by Dub 2 · 0⤊ 0⤋