Respuesta :
Answer : 42 grams of oxygen are required.
Explanation :
Step 1 : Write balanced chemical equation.
The combustion reaction of octane with oxygen can be written as,
[tex] 2 C_{8}H{18} (g) + 25 O_{2} (g) \rightarrow 16 CO_{2} (g) +18 H_{2}O (g) [/tex]
Step 2 : Find moles of octane.
The given mass of octane is 12 g
Molar mass of octane is 114.2 g/mol
The moles of octane are calculated as,
[tex] moles =\frac{grams}{MolarMass} =\frac{12g}{114.2g/mol} = 0.105 mol [/tex]
We have 0.105 mol octane.
Step 3: Use mole ratio from balanced equation to find moles of O₂
The mole ratio of octane and O₂ is 2 : 25. Let us use this as a conversion factor.
[tex] 0.105 mol C_{8}H_{18} \times\frac{25mol O_{2}}{2molC_{8}H_{18}} = 1.31 mol [/tex]
We have 1.31 mol O₂.
Step 4 : Convert moles of O₂ to grams.
The molar mass of O₂ is 32 g/mol.
The grams of O₂ can be calculated as,
[tex] 1.31 mol O_{2} \times\frac{32g}{mol} = 42 grams [/tex]
42 grams of oxygen are required.
