When 25.6 g [tex]C_{8} H_{18}[/tex] is burned, the moles of [tex]CO_2[/tex] emitted into the atmosphere is 1.80 mol
Explanation:
The molar mass of octane [tex]C_{8} H_{18}[/tex] is calculated as 8(12) +18(1) = 114 g/mol.
Where atomic mass of Carbon is C=12, and Hydrogen H=1.The sample contains 25.6 g of octane, Therefore
[tex]\frac{25.6 g}{114 g / m o l}=0.225 \mathrm{mol}[/tex]
The combustion of octane can be written as:
[tex]2 C_{8} H_{18}+25 O_{2} \rightarrow 16 C O_{2}+18 H O_{2}[/tex]
The mole ratio can be represented as:
[tex]C O_{2}: C_{8} H_{18}[/tex] = 16:2 = 8:1.
Hence, the quantity of carbon dioxide emitted is:
[tex]\frac{8}{1} \times 0.225 \mathrm{mol}[/tex] = 1.80 mol