Balanced equation of H2O and C9H8O4?

Answer 1

2 C8H18 + 25 O2 = 16 CO2 + 18 H20

Do it by saying C8H18 + a O2 = m CO2 + n H2O. Then match up all the atoms so you get for carbon 8=m, hydrogen 18 = 2n, oxygen 2a = 2m+n. This gives you m=8, n=9 and a=25/2. Since we don't like fractions in these equations we can sort that out by just doubling the whole way through to get the quoted result given at the beginning.

Answer 2

2C8H18 + 25O2 ----> 16CO2 + 18H2O

The easiest way to solve this is to start with balancing the carbons. The molecule on the left has 8 carbons, so you will need 8 carbon dioxide molecules on the right. Next look at your hydrogens - you have 18 on the left, and on the right the hydrogens are in sets of two in the water molecule. However, putting a coefficient of 9 in front of the H2O means you have an odd number of oxygens, which you can't balance on the left. Therefore, double all the coefficients you have so far (16 instead of 8 CO2, 2 instead of one C8H18, 18 instead of 9 H2O). This keeps the carbons and hydrogens balanced from before, and now you have 50 oxygens on the right. Divide this by 2 (since elemental oxygen is diatomic, as shown by the subscript on the left) and you have a coefficient of 25 for your oxygen.

Answer 3

The way to do this is to use variables for the amount of each type of atom:

a C8H18 + b O2 = c CO2 + d H2O

You want to solve for a, b, c, and d.

Now, write an equation for each type of atom. In this case:

For C: 8a = c

For H: 18a = 2d

For O: 2b = 2c + d

I am leaving the fun part for you - to solve the simultaneous equations. Note: You have 4 variables, but only 3 equations. By solving, you will have relationships between the amounts, but no absolute amounts. This is logical, for you can (for example) double each quantity, and the equation will still be valid. Adjust the final result so that you get integers. For example, if you get 1/2 of one type of molecule, and 1/3 of another, multiply by the common denominator (6 in this example).

Answer 4

C8H18 means mix oxygen with octane and that's the formular

so the equation is:

Octane+Oxygen-------> carbon dioxide + water (combustion)

Answer 5

2 c8h18 + 25 o2 ---> 16 co2 + 18 h2o

Answer 6

c3h8+5o2=3co2+4h2o

Answer 7

c5h12 + 8o2 -----> 5co2 + 6h2o

Answer 8

Ch3cooh+c7h6o3

Answer 9

2 C8H18 + 17 O2 = 16 CO + 18 H2O

Answer 10

C5H12+O2=CO+H2O