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

1 Answer
May 6, 2017

Answer:

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#.