Question

Question #86eb9

Answer 1

All hydrocarbons combust to give carbon dioxide and water.......

Explanation:

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)`

`C_4H_10(g) + 13/2O_2(g) rarr 4CO_2(g) + 5H_2O(l)`

And if combust a `40*g` mass of propane......you gets.....

`(40*g)/(44.1*g*mol^-1)xx3xx44.01*g*mol^-1=119.8*g*CO_2`

And if combust a `60*g` mass of butane......you gets.....

`(60*g)/(58.12*g*mol^-1)xx4xx44.01*g*mol^-1=181.7*g*CO_2`