Stoichiometry 2 For the given combustion reaction of octane

Stoichiometry 2 For the given combustion reaction of octane, CgH18, answer the following questions: (Answers to the questions are given in parenthesis.) 2CgH18 250216CO2 18H20 a. Write 5 possible molar ratios from this equation. b. How many moles of Co2 would be produced by reacting 0.67 moles of octane with excess of oxygen? (Amount of oxygen is not involved in the calculation) (5.4 mol CO2) How many males of H20 would be produced by reacting 0.67 moles of octane with excess of oxygen?

Solution

Part a

Five molar ratios are

O2/C8H18 = 25/2 = 12.5

CO2/C8H18 = 16/2 = 8

H2O/C8H18 = 18/2 = 9

CO2/O2 = 16/25 = 0. 64

H2O/O2 = 18/25 = 0.72

Part b.

From the stoichiometry of the reaction

Excess reactant = O2

Limiting reactant = C8H18

2 mol of C8H18 produces = 16 mol of CO2

0.67 mol of C8H18 produces = 16*0.67/2 = 5.36 mol of CO2

Moles of CO2 formed = 5.4 mol

Part C

From the stoichiometry of the reaction

Excess reactant = O2

Limiting reactant = C8H18

2 mol of C8H18 produces = 18 mol of H2O

0.67 mol of C8H18 produces = 18*0.67/2 = 6.03 mol of H2O

Moles of H2O formed = 6 mol

Part d

Moles of C8H18 = mass/molecular weight

= 225g / (8*12+18*1) g/mol

= 1.9736 mol

From the stoichiometry of the reaction

2 mol of C8H18 reacts with = 25 mol of O2

1.9736 mol of C8H18 reacts with = 25*1.9736/2 = 24.67 mol of O2

Moles of O2 required = 24.7 mol

Part e

From the stoichiometry of the reaction

Excess reactant = O2

Limiting reactant = C8H18

2 mol of C8H18 produces = 16 mol of CO2

1.9736 mol of C8H18 produces = 16*1.9736/2 = 15.78 mol of CO2

Moles of CO2 formed = 15.8 mol