provided the combustion reaction C3H8+5O2->3CO2+4H2O, how may liters the carbon dixide can be created from 1.00 liters the O2 at conventional temperature and also pressure?
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You are watching: C3h8+5o2=3co2+4h2o `C_3H_8+5O_2 rarr 3CO_2+4H_2O`

At STP 1 mol of `O_2` has 22.4L.

Therefore;

1L of `O_2` consists of `1/22.4` mol (0.045mol)

So we have actually 0.045mol that `O_2` .

Mole ratio

`O_2:CO_2 = 5:3`

Amount of `CO_2` formed `= 0.045/5xx3 = 0.027mol`

at STP 1mol of `CO_2` contains...

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`C_3H_8+5O_2 rarr 3CO_2+4H_2O`

At STP 1 mol of `O_2` contains 22.4L.

Therefore;

1L the `O_2` includes `1/22.4` mol (0.045mol)

So we have 0.045mol of `O_2` .

Mole ratio

`O_2:CO_2 = 5:3`

Amount the `CO_2` created `= 0.045/5xx3 = 0.027mol`

At STP 1mol the `CO_2` consists of 22.4L.

Therefore;

Volume the `CO_2` formed `= 22.4/1xx0.027 = 0.6L`

So 0.6L that `CO_2` will certainly be developed when 1L of `O_2` reacted v `C_3H_8` .

Assumption

`CO_2` and `O_2` behaved as ideal gasses.

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