Answer
983 L
Work Step by Step
2C8H18 + 25O2 -----------> 16CO2(g) + 18H2O(g)
balanced equation shows that
2 moles C8H18 produces 16 mole CO2 and 18 moles H2O (a total of 34 moles gas)
So 1 mole C8H18 produces a total of 17 moles of gas
moles C8H18 = mass / molar mass
= 125 g / 114.224 g/mol
= 1.094 moles
Therefore moles gas products = 17 x 1.094 mol
= 18.6 moles of gaseous products.
V = 18.6 mol x 62.36367 torrLmol^-1K^-1 x 623.15 K / 735 torr
= 983 L
b) 2 moles C8H18 requires 25 moles of O2 to fully react
So 1.094 moles C8H18 needs (25/2 x 1.094) moles O2
= 13.675 moles O2 needed
Volume of this many moles of O2 under the given conditions is
V = 13.675 mol x 62.36367 x 623.15 K / 735 torr
= 723 L
O2 is 21.0 % of air, thus 723 L is 21.0 % of the total volume of air needed
21.0 / 100 x volume air = 723 L
Therefore volume air needed = 723 L / 0.21
= 3443 L
Air in exhaust = total air - O2 consumed = 3443 L - 723 L = 2720 L
Total gases in exhaust = unused air + gaseous products
= 2720 L + 983 L
= 3703 L
