Respuesta :

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

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