Sure, let's break down the problem step-by-step to determine how many moles of [tex]\( CO_2 \)[/tex] are emitted when 13.1 grams of octane ([tex]\( C_8H_{18} \)[/tex]) is burned.
### Step 1: Determine the molar mass of octane ([tex]\( C_8H_{18} \)[/tex])
The molar mass of octane is calculated by adding the atomic masses of all the atoms in the molecule:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
Thus, for octane ([tex]\( C_8H_{18} \)[/tex]):
[tex]\[ \text{Molar mass of } C_8H_{18} = (8 \times 12.01 \, \text{g/mol}) + (18 \times 1.01 \, \text{g/mol}) \][/tex]
[tex]\[ \text{Molar mass of } C_8H_{18} = 96.08 \, \text{g/mol} + 18.18 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } C_8H_{18} = 114.26 \, \text{g/mol} \][/tex]
### Step 2: Calculate the number of moles of octane burned
Given mass of octane is 13.1 grams. The number of moles of octane is found using the formula:
[tex]\[ \text{Moles of octane} = \frac{\text{Mass of octane}}{\text{Molar mass of octane}} \][/tex]
[tex]\[ \text{Moles of octane} = \frac{13.1 \, \text{g}}{114.26 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of octane} \approx 0.1147 \, \text{mol} \][/tex]
### Step 3: Use stoichiometry to find the moles of [tex]\( CO_2 \)[/tex] produced
From the balanced chemical equation:
[tex]\[ 2 C_8H_{18} + 25 O_2 \rightarrow 16 CO_2 + 18 H_2O \][/tex]
The stoichiometric ratio shows that 2 moles of [tex]\( C_8H_{18} \)[/tex] produce 16 moles of [tex]\( CO_2 \)[/tex]. Therefore, 1 mole of [tex]\( C_8H_{18} \)[/tex] produces 8 moles of [tex]\( CO_2 \)[/tex].
[tex]\[ \text{Moles of } CO_2 \text{ produced} = \text{Moles of octane} \times 8 \][/tex]
[tex]\[ \text{Moles of } CO_2 \text{ produced} \approx 0.1147 \, \text{mol} \times 8 \][/tex]
[tex]\[ \text{Moles of } CO_2 \text{ produced} \approx 0.9174 \, \text{mol} \][/tex]
### Final Answer
The number of moles of [tex]\( CO_2 \)[/tex] emitted into the atmosphere when 13.1 grams of [tex]\( C_8H_{18} \)[/tex] is burned is approximately 0.9174 moles.
