Question

Please solve the q 92?

Answer 1

`C_5H_12`

Explanation:

So we know how much product we got from the reaction between hydrocarbon (HC) and oxygen.

`CO_2 = 2500 mL`
`H_2O = 3000 mL`

And we used
`HC = 500 mL`

We do not know how many hydrogens and carbons the HC consisted of, therefore we can write the reaction happening and then finding the amounts. Formula for HC is `C_xH_y` (Since we dont know them):

`C_xH_y + (x+y/4)O_2 -> x*CO_2+ y/2 * H_2O`

This gives:

`500 mL + 500 mL (x+y/4) = 500 mL*x + 500 * y/2`

We need to look at the right side of the equation. Here we can see that if x is `CO_2` and y is `H_2O`, we would have the following equations for x and y:

`500 mL *x=2500 mL -> x=(2500 mL)/(500mL) = 5`

`500 mL* y/2 = 3000 mL -> y= (3000 mL)/(250 mL) = 12`

Thereby we have
`C_5H_12`