1 Answer | Add Yours
`C_3H_8+5O_2 rarr 3CO_2+4H_2O`
At STP 1 mol of `O_2` contains 22.4L.
1L of `O_2` contains `1/22.4` mol (0.045mol)
So we have 0.045mol of `O_2` .
`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.
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` .
- `CO_2` and `O_2` behaved as ideal gasses.
We’ve answered 318,989 questions. We can answer yours, too.Ask a question