Question

A sample of liquid heptane (C₇H₁₆) weighing 11.50g is reacted with 1.300 moles of oxygen gas. The heptane is burned completely. After the reaction is complete, the amount of gas present is 1.050 moles.

a. How many moles of CO are produced?

b. How many moles of CO₂ are produced?

c. How many moles of O₂ are left over?

Answer 1

C7H16 + 11 O2 ------------------> 7 CO2 + 8 H2O

    1              11                                7                8

0.115          1.30                              ?

here limiting reagent is heptane . so product formed based on that.

1 mol of heptane -------------- 7 mol of CO2

0.115 mol of heptane ------------ ?? CO2

moles of CO2 produced = 0.115 x 7 = 0.805 moles

moles of H2O produced = 0.115 x 8 = 0.92 moles

here limiting reagent is heptane . so heptane is consumed completely. O2 is remained. that is

amount of O2 required = 0.115 x 11 = 1.265

we have 1.30 moles

amount of O2 remained = 1.30 - 1.265 = 0.035 moles

moles of O₂ are left over = 0.035 moles