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3 years ago

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Consider the combustion of octane (C8H18)

1 answer:

Masja [62]3 years ago

8 0

First, you have to find now many moles of octane are present in 191.6g of octane. To do this you need to do this you need to divide 191.6g by its molar mass (which is 114g/mol). This will give you 1.681 moles of octane. Then you need to use the fact that 2 moles of octane are us ed to make 16 moles of carbon dioxide to find how many moles of carbon dioxide 1.681mole of octane produces. To do this you need to multiply 1.681mole by 16/2 to get 13.45mol carbon dioxide. The final step is to find the number of grams presswnt in 13.446 moles of carbon dioxide. To do this you need to multiply 13.446 mole by carbon dioxides molar mass (which is 44g/mol) to get 591.6 g of carbon dioxide.
Therefore, 591.6g of carbon dioxide is produced when 191.6 grams of octane is burned.

I hope this helps. Let me know in the comments if anything is unclear.

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V is the volume of the gas in L (V = 20.0 L).

n is the no. of moles of the gas in mol (n = ??? mol).

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To find the mass of N₂ in the cylinder, we can use the relation:

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