Question

Liquid hexane ( CH3(CH2)4CH3 ) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 5.17 g of hexane is mixed with 7.5 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.

Answer 1

Reaction eqn is

2 C6H14 + 19 O2 --> 12 CO2 + 14 H2O

5.17 g of hexane= 5.17/86 moles = 0.06 moles hexane

7.5 g O2 = 7.5/16 = .468 moles

by stoichiometry a mole hexane needs 9.5 moles O2 for complete combustion

0.06 moles hexane needs-> 0.06 *9.5 moles O2 = .57 moles of O2

but O2 moles are insufficient hence it is our limiting reactant

.468 moles of O2 means -> (.468 * 12/19) moles of CO2 by stoichiometry

= .29 moles of CO2

mass of CO2 -> .2955 * 44 = 12.76 grmas of CO2