Balancing Equations?

Question

i can't get this one can anyone help

__C10H16(OH)3 +__O2 = __CO2 +__H2O

Answer 1

First look at C and declare some variables (X,Y,Z)
(look at C first because it only appears once on each side)

X C10H16(OH)3 + Z O2 = 10X CO2 + Y H2O

Now we can forget about C because it's already ok.

For H:
19X = 2Y

For O:
3 X + 2Z = 20X + Y

So we solve these equations and get:
Y = 19/2 X
2Z = 17X + Y = 34/2 X + 19/2 X = 53/2 X
Z = 53/4 X

We need whole numbers so X = 4
Y = 38 and Z = 53

Checking this gives:
4*10 C = 40 C ok
4*16 + 4*3 H = 2*38 H ok
4*3 + 53*2 O = 40*2 O + 38 O ok

Answer 2

C10H16(OH)3 has 10 carbons, three oxygens and 19 hydrogens.
We know that all the carbon forms CO2, so we'll need to form 10 molecules of CO2 for every molecule of C10H16(OH)3. If we form 10 molecules of CO2 we need twenty oxygens in the reactants, and we've got 3, which means we'll need 17 more. We'll need even more oxygens in the reactants, because H20 is also a product.

Now look at hydrogens. All 19 hydrogens in C10H16(OH)3 form 19/2 molecules of H2O. This uses up 19/2 oxygen. So we need 17 + 19/2 = 53/2 additional Oxygens. The source for those 53/2 oxygens is two oxygens per O2, so we need 53/4 O2.

C10H16(OH)3 + 53/4O2 = 10CO2 + 19/2H2O is the basic equation, but as you may have noticed, it has a fractional molecules. The smallest fraction is in fourths, so this suggests we quadruple the reaction. This can be done by mutiplying each component by 4.

So, 4*C10H16(OH)3 + 4*53/4O2 = 4*10CO2 + 4*19/2H2O
4C10H16(OH)3 +53O2 = 40CO2 + 38H2O

Answer 3

4+53=40+38

Answer 4

2 C10H16(OH)3 + 28 O2 --> 20 CO2 + 19 H2O

Answer 5

4...53...= 40...38

Tip: Its ok to get fraction...juz multiply by a common denominator in the end...

Answer 6

4C10H16(OH)3 + 53O2 -> 40CO2 + 38 H2O

Answer 7

4C10H16(OH)3 + 53O2 -> 40CO2 + 38 H2O

I hope this helps!!

Answer 8

4
53
40
38