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8090 [49]
3 years ago
14

Consider the combustion of octane (C8H18)

Chemistry
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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2. A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is the balloon's volume
Sphinxa [80]

Answer:

3.01 L

Explanation:

V

1

: 2.75L

T

1

:  

18

∘

C

V

2

: ?

T

2

:  

45

∘

C

If you know your gas laws, you have to utilize a certain gas law called Charles' Law:

V

1

T

1

=

V

2

T

2

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1

is the initial volume,  

T

1

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V

2

is final volume,  

T

2

is final temperature.

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

2.75

291

=

V

2

318

By cross multiplying, we have...

291

V

2

= 874.5

Dividing both sides by 291 to isolate  

V

2

, we get...

V

2

= 3.005...

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

Therefore,  

V

2

= 3.01 L

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