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Balance the following chemical equation, then answer the following question.
C8H18(g) + O2(g) →CO2(g)+H2O(g)
How many grams of oxygen are required to react with 20.0 grams of octane (C8H18) in the combustion of octane in gasoline?
Express the mass in grams to one decimal place.

Respuesta :

manselandy
The balanced equation is as follows:
2C8H18(g) + 25O2(g) →16CO2(g)+18H2O(g)
The number of moles in 20 grams of octane can be calculated as follows:
No. Of moles=mass/relative molecular mass.
=20/(12*8+1*8)
=20/104
=0.2moles
The reaction ratio is 2:25
Hence if 2 moles octane react with 25 moles of oxygen
Then the number of moles of oxygen reacting with 0.2 moles of octane is calculated as follows :
(0.2*25)/2=2.5 moles
2.5 moles of oxygen have a mass of:
2.5moles* 32(the molecular mass of oxygen)=800.0grams

Answer: 70.2 g

Explanation:

[tex]2C_8H_{18}(g)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]

According to the law of conservation of mass, the mass of the products must be same as the mass of the reactants in every chemical equation. Thus the number of atoms of every element must be same on both sides of the equation.

2 moles of octane weigh =[tex]2\times 114g/mol=228g[/tex]

25 moles of oxygen weigh =[tex]25\times 32g/mol=800g[/tex]

Thus 228 g of octane reacts with 800 g of Oxygen.

20 g of octane reacts with [tex]=\frac{800}{228}\times 20=70.2[/tex] g of Oxygen.



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