Question

A sample of liquid heptane (C7H16) weighing 11.5 g is reacted with 1.3 mol of oxygen gas. The heptane is burned completely (heptane reacts with oxygen to form both carbon monoxide and water and carbon dioxide and water). After the reaction is complete, the amount of gas present is 1.050mol (assume that all of the water formed is liquid).

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

a) 0.525 mol

b) 0.525 mol

c) 0.236 mol

Step-by-step explanation:

The combustion reactions (partial and total) will be:

C₇H₁₆ + (15/2)O₂ → 7CO + 8H₂O

C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O

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2C₇H₁₆ + (37/2)O₂ → 7CO + 7CO₂ + 16H₂O

It means that the reaction will form 50% of each gas.

a) 0.525 mol of CO

b) 0.525 mol of CO₂

c) The molar mass of heptane is: 7*12 g/mol of C + 16*1 g/mol of H = 100 g/mol

So, the number of moles is the mass divided by the molar mass:

n = 11.5/100 = 0.115 mol

For the stoichiometry:

2 mol of C₇H₁₆ -------------- (37/2) mol of O₂

0.115 mol of C₇H₁₆ --------- x

By a simple direct three rule:

2x = 2.1275

x = 1.064 mol of O₂

Which is the moles of oxygen that reacts, so are leftover:

1.3 - 1.064 = 0.236 mol of O₂