Q3.121CP

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. (a) How many grams of oxygen are needed for complete combustion of the toluene? (b) How many total moles of gaseous products form? (c) How many molecules of water vapor form?

Step-by-Step Solution

Verified
Answer

a. To complete the combustion of the toluene, 54.1 g of oxygen is needed.

b. A total of 2.07 moles of gaseous products forms.

c. A total of 4.53×1023molecules of water vapor forms.

1Step 1: Calculating the oxygen needed for the complete combustion of toluene

a.

Multiply the given toluene volume and the molar mass reciprocal to find the mole number of the toluene.

MoleofC7H8=20mL×0.867gC7H81mLC7H8×1molC7H892.14gC7H8=0.188molC7H8.MassofO2needed=0.188molC7H8×9molO21molC7H8×32gO21molO2=51.1gO2.

2Step 2: Calculating the total moles of gaseous products formed

b. Calculate the total moles of gaseous products as shown below:

MolesofCO2formed=0.188molC7H8×7molCO21molC7H8=1.316molCO2.MolesofH2Oformed=0.188molC7H8×4molH2O1molC7H8=0.752molCO2.Totalmole=1.316molCO2+0.752molH2O=2.07mol.

3Step 3: Calculating the water molecule formation

c. Calculate the number of water molecules as shown below.

NumberofH2Omolecules=0.752molH2O×6.022×1023H2Omolecules1molH2O=4.53×1023H2Omolecules.