Question

2C31H64 + 63O2 --> 62CO2 + 64H2O Whatmass of water will be produced in the reaction? Mass before burning 0.72 g after burning 0.27 g

Answer 1

We write first the stoichiometric equation.........

Explanation:

`C_31H_64(s) + 47O_2(g) rarr 31CO_2(g) + 32H_2O(l)`

Now `C_31H_64` is probably a wax or even a tar......Practically, we would expect incomplete combustion with such a hydrocarbon.

If we assume complete combustion, then.............(and I assume, perhaps wrongly, that a `0.5*g` mass combusts)

`"Moles of hydrocarbon"=(0.50*g)/(436.85*g*mol^-1)=1.14xx10^-3*mol`.

And the equation CLEARLY indicates that `1.14xx10^-3*molxx32=3.66xx10^-2*mol` water result, i.e. a mass of `0.659*g`.