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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.

1 Answer

6 votes

Answer: 70.2 g

Explanation:


2C_8H_(18)(g)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)

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 =
2* 114g/mol=228g

25 moles of oxygen weigh =
25* 32g/mol=800g

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

20 g of octane reacts with
=(800)/(228)* 20=70.2 g of Oxygen.



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