Solution Found!
One of the compounds used to increase the octane rating of
Chapter 3, Problem 83P(choose chapter or problem)
One of the compounds used to increase the octane rating of gasoline is toluene (right). Suppose \(20.0 mL\) of toluene (d \(0.867 g/mL\)) is consumed when a sample of gasoline burns in air.
How many grams of oxygen are needed for complete combustion of the toluene? How many total moles of gaseous products form? How many molecules of water vapor form?
Equation Transcription:
Text Transcription:
20.0 mL
0.867 g/mL
Questions & Answers
QUESTION:
One of the compounds used to increase the octane rating of gasoline is toluene (right). Suppose \(20.0 mL\) of toluene (d \(0.867 g/mL\)) is consumed when a sample of gasoline burns in air.
How many grams of oxygen are needed for complete combustion of the toluene? How many total moles of gaseous products form? How many molecules of water vapor form?
Equation Transcription:
Text Transcription:
20.0 mL
0.867 g/mL
ANSWER:
Solution 83P
Step 1
(a) Here we have to calculate how many grams of oxygen are needed for complete combustion of the toluene.
The balanced chemical equation for the complete combustion of toluene is,
C7H8 (l) + 9 O2 (g) → 7CO2 (g) + 4 H2O (g)
In this question it has been given that the volume of toluene is 20.0 mL and density is 0.867 g/mL. Then we can calculate the mass of toluene as, mass = density volume
= 0.867 g/mL20.0 mL
= 17.34 g
From the above balanced equation it has been found that, 9 moles of O2 has combined with 1 mole of C7H8.
Thus, the moles of toluene = 17.34 g
=0.188 mol
Hence the mass of oxygen can be calculated as,
= 0.188 mol of C7H8
= 54.14 g of O2
Thus 54.14 grams of oxygen are needed for complete combustion of the toluene.