How do you solve word equation C6H14 plus O2----CO2 plus H20?

Answer 1

Include constants x,y,z as in xC6H14 + yO2 = zCO2 + H2O (notice the number of constants is 1 less than the number of terms). Then equate numbers of like atoms on each side, because atoms are not created or destroyed. So for Oxygen you get 2y=2z+1 and using the other atoms you'll get more equations. Solve these as a set of simultaneous equations. If you get fractions in the answers, multiply all by one same number (ie equally) to get rid of them. That's how.

Answer 2

To balance this equation, you must ensure the number of each type of atom is the same on both sides by adjusting coefficients. The balanced equation is C6H14 + 9O2 -> 6CO2 + 7H2O. This means that 1 molecule of C6H14 reacts with 9 molecules of O2 to produce 6 molecules of CO2 and 7 molecules of H2O.

Answer 3

You have to think a bit 'obtusely/backwardly'.

C6H14 + O2 = CO2 + H2O

We note that the alkane has '6' carbons , so we can put a '6' in front of the CO2

Similarly, there are '14' hydrogens , so we put a '7' in front of the 'H2O' . There being already two atoms of hydrogen in each molecule of water , so 7 x 2 = 14

Hence

C6H14 + O2 = 6CO2 + 7H2O

This is where we think obtusely. We note that there are 6 x 2 = 12 oxygens in CO2 , and 7 x 1 = 7 oxygens in H2O.

So on the product side we have 12 + 7 = 19 oxygens.

So on the reactant side the 'O2' has to be made up to '19' oxygens. This is done by placing a molar ratio of 19 = 9/2 on the reacting oxygens.

Hence

C6H14 + 19O2 = 6CO2 + 7H2O

This gives us '38' oxygens. So to balance the other three components are multiplied by '2'

Hence

2C6H14 + 19O2 = 12CO2 + 14H2O

Balanced.

You will notice

There are

2 x 6 = 12 (C)

2 x 14 = 14 x 2 =28(H)

19 x 2 = (12 x 2)+ (14 x 1) = 38 (O)

NB In working out these organic combustion problems. Initially balance the numbers of 'C' & 'H' on the product side. Then look at the number of oxygens on the product side, and working back to the number of oxygens required to react.