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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?

Short Answer

Expert verified

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.

Step by step solution

01

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.

Mole of C7H8=20 mL×0.867 g C7H81 mL C7H8×1 mol C7H892.14 g C7H8                                           =0.188 mol C7H8.Mass of O2 needed=0.188 mol C7H8×9 mol O21 mol C7H8×32 g O21 mol O2                                             =51.1 g O2.

02

 Step 2: Calculating the total moles of gaseous products formed

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

Moles of CO2 formed=0.188 mol C7H8×7 mol CO21 mol C7H8                                                  =1.316 mol CO2.Moles of H2O formed=0.188 mol C7H8×4 mol H2O1 mol C7H8                                                  =0.752 mol CO2.                         Total mole=1.316 mol CO2+0.752 mol H2O                                                  =2.07 mol.

03

 Step 3: Calculating the water molecule formation

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

Number of H2O molecules=0.752 mol H2O×6.022×1023H2Omolecules1 mol H2O                                                            =4.53×1023H2O molecules.

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