If 2.0 moles of octane vapor, C_8H_18, is reacted with 15 moles of oxygen to pro

Question

If 2.0 moles of octane vapor, C_8H_18, is reacted with 15 moles of oxygen to produce carbon dioxide and water, what are the partial pressures of all gases in the mixture at a total pressure of 1 atm? Propane (C_3Hg) gas is used in gas grills and is combusted with oxygen to produce carbon dioxide and water vapor. If 5.00 L of propane is combusted, what volume of oxygen is required and what volumes of water vapor and carbon dioxide are produced? Calcium carbonate (25.00 grams) is reacted with hydrochloric acid to produce calcium chloride, water, and carbon dioxide gas at 0.987 aim and 25 degree C. If the carbon dioxide gas collected over water at 25 degree C what is the volume of dry carbon dioxide at standard conditions?

Explanation / Answer

15)
the reaction taking place is:
2 C8H18 + 25 O2 ---> 16 CO2 +18 H2O.

2 mol of octane needs 25 mol of O2
But we have 15 moles of O2
So, O2 is limiting reagent

whole O2 will raect
mole of CO2 formed = (16/25)*15 = 9.6 mol
mole of H2O formed = (18/25)*15 = 10.8 mol
mol of octane remaining = 2 - (2/25)*15 = 0.8 mol
mol of O2 = 0

total number of mol = 9.6 + 10.8 + 0.8 = 21.2 mol

Poctane = (number of mol of octane) * total P / total mol
= 0.8*1/21.2
=0.038 atm

similarly Po2 = 0
P CO2 = 9.6*1/21.2 = 0.453 atm

rest of pressure will be due to water
p H2O = 1-0.038 - 0.453 = 0.512 atm

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