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

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

Therefore;

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

So we have 0.045mol of `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 22.4L.

Therefore;

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

*So 0.6L of `CO_2` will be formed when 1L of `O_2` reacted with
`C_3H_8` .*

*Assumption*

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

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