The combustion of octane, c8h18, proceeds according to the reaction 2C8H18(l) +
ID: 899910 • Letter: T
Question
The combustion of octane, c8h18, proceeds according to the reaction 2C8H18(l) + 2502(g) rightarrow 16C02(g)+18H_2o(l) if 321 mol of octane combusts. what volume of carbon dioxide Is produced at 26. 0degrees and 0. 995 atm? A sample of oxygen gas was collected via water displacement Since the oxygen was collected via water displacement, trie sample is saturated with water vapor If the total pressure of the mixture at 26. 4degrees is 809 torr. what Is the partial pressure of oxygen? The vapor pressure of water at 26. 4 degree Is 25. 81 mm Hg. A 9. 70-L container holds a mixture of two gases at 11degrees. The partial pressures of gas A and gas B. respectively, are 0. 416 atm and 0. 587 atm. If 0. 210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become? 3. 95 g of an unknown gas at 29degrees and 1. 10 atm Is stored m a 2. 15-L flask. What is the density of the gas?Explanation / Answer
Solution 1
2 mole of Octane gives 16CO2
Therefore 321 moles of octane gives 321*16/2 = 2568 moles
Applying , PV = nRT
= 0.995 * V = 2568 * 0.0821 * 299
= 63355.78 L
Solution 2
The partial pressure of Oxygen = 809-25.81 = 783.19 torr
Solution 3
PV = nRT
= 9.7 * (0.416+.587) = moles * 0.0821*284
= moles = 0.417
Now we add 0.21moles
Applying pV = nRT
= p* 9.7 = 0.627*0.0821*284
= pressure = 1.507 atm
Solution 4
pV = nRT
= 1.1*2.15 = moles * 0.0821*302
= moles = 0.09538
Now mass/Ml wt = 0.09538
= ml wt = 3.95/0.09538
= 41.411 grams
PM = dRT
= 1.1*41.411 = d*0.0821*302
= 1.837 = density