Given the following unbalanced equation for the combustion of octane, caculate the grams of O2 consumed by the

Question

reaction with 858 g C8H18?

C8H18 + O2----------CO2 + H2O

For that reaction in that problem, if 60.0 kg of C8H18 is completely combusted, how many kg of CO2 and H2O are produced, by converting kg to grams?

Answer 1

To start with, you have to balance the equation. Every mole of octane requires 12.5 moles of oxygen, and gives off 8 moles of CO2 and 9 moles of water.

Then you have to work out the Mr of octane.

(858/Mr) x 12.5 x 32 is your answer.

For the other problems, you must work in grams. You can convert back to kg later.

Answer 2

2CH4 + 3O2 = 2CO2 + 2H2O + 2H2 whilst CH4 react with O2 decrease than extreme temperature and stress and nickel catalyst,it truly is going to transform in to CO2,H2O and H2.decrease than frequent circumstance this reaction not conceivable.