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?

Short 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.

Step-by-step solution

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.

$\begin{array}{rcl}Mole of C_7{H}_8& =& 20 mL\times \frac{0.867 g C_7{H}_8}{1 mL C_7{H}_8}\times \frac{1 mol C_7{H}_8}{92.14 g C_7{H}_8}\\ & =& 0.188 mol C_7{H}_8.\\ Mass of O_2 needed& =& 0.188 mol C_7{H}_8\times \frac{9 mol O_2}{1 mol C_7{H}_8}\times \frac{32 g O_2}{1 mol O_2}\\ & =& 51.1 g O_2.\end{array}$

 Step 2: Calculating the total moles of gaseous products formed

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

$\begin{array}{rcl}Moles of C{O}_2 formed& =& 0.188 mol C_7{H}_8\times \frac{7 mol C{O}_2}{1 mol C_7{H}_8}\\ & =& 1.316 mol C{O}_2.\\ Moles of H_{2}O formed& =& 0.188 mol C_7{H}_8\times \frac{4 mol H_{2}O}{1 mol C_7{H}_8}\\ & =& 0.752 mol C{O}_2.\\ Total mole& =& 1.316 mol C{O}_2+0.752 mol H_{2}O\\ & =& 2.07 mol.\end{array}$

 Step 3: Calculating the water molecule formation

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

$\begin{array}{rcl}Number of H_{2}O molecules& =& 0.752 mol H_{2}O\times \frac{6.022\times {10}^{23}{H}_{2}Omolecules}{1 mol H_{2}O}\\ & =& 4.53\times {10}^{23}{H}_{2}O molecules.\end{array}$