Balancing Equation help!

[Laila]

I need to balance this equation...and the Hydrogens are kinda throwing me off with balancing the number of atoms for it

__HC2H3O2 + __Ba(OH)2 > __H2O + __Ba (C2H3O2)2

thanks in advance.

[Admin Khan]

Okay, this is a simple acid/bace equation.
The "H" from the Acetic Acid will combine with the "OH" from your base to produce H20.
You can use the formula for double displacement (Ax +By -> Bx + Ay) to combine the Acetate with the Barium.

So, you have:
HC2H3O2 + Ba(OH)2 -> H2O + Ba(C2H3O2)2

Since Barium has a charge of +2 and Acetate has a charge of -1, you need 2 Acetate compounds to balance their charge.

Now, you need to count the elements on each side of the equation, to make sure they are balanced.
There are 6 H atoms on the left side, but there are 8 H atoms on the right side.
There are 2 atoms of C on the left side of the equation, but there are 4 on the right.
There are 4 atoms of O on the left side, but there are 5 on the right side.
And there are 2 atoms of Ba on both sides.

So, the H, C, and O need to be balanced.

By placing coefficients in front of compounds in the formula, you can easily balance the equation.

Balanced equation:
2 HC2H3O2 + Ba(OH)2 -> 2 H20 + Ba(C2H3O2)2


To check, again count the number of elements on each side of the equation.

There are 10 H atoms on both sides, 4 C atoms on both sides, 6 O atoms on both sides, and 2 Ba atoms on both sides.

The chemical equation is balanced.

[Laila]

^ thank you

[Baidariwal]

I need to balance this equation...and the Hydrogens are kinda throwing me off with balancing the number of atoms for it

__HC2H3O2 + __Ba(OH)2 > __H2O + __Ba (C2H3O2)2

thanks in advance.

2HC2H3O2 + Ba(OH)2 --> 2H2O + Ba (C2H3O2)2

[Baidariwal]

oops too late again :/

[Laila]

Thanks i got it now...i knew i had it, i just forgot to add my 2H2O for the product since its on the computer....thanks

[Admin Khan]

oops too late again :/

Your effort is appreciated Baidariwal. You might even help me when I am in need of assistance. Never feel discouraged wror jaana.

[Haider Bangash 2.0]

I know this is late but let me just throw my two cents for those who might need it later. The algebraic method of balancing chemical equations that I am gonna show is good for balancing more complicated reactions.

HC2H3O2 + Ba(OH)2 -> H2O + Ba(C2H3O2)2

What we do is place letters in front of each molecule:
aHC2H3O2 + bBa(OH)2 -> cH2O + dBa(C2H3O2)2

Now we write # of each atom in terms of these letters (multiply the letter by the # of each atom), like so:

Let's solve in terms of a (you can solve for any variable that you like)
Carbon:
2a = 4d
d = (1/2)a

Barium:
b = d

Oxygen
2a + 2b = c + 4d
Since d = (1/2) from above equation we rewrite this as:
2a + 2b = c + 4(1/2)a
2a + 2b = c + 2a
c = 2d
Since b = d (see Barium), and d = (1/2)a:
c = 2(1/2)a
c = a

Now that we have solved all the variables in terms of a, we can rewrite the reaction as:
a = a
b = d = (1/2)a
c = a
d = (1/2)a

aHC2H3O2 + (1/2)aBa(OH)2 -> aH2O + (1/2)Ba(C2H3O2)2
Multiply by 2 to get rid of fractions (since you can't have "half" of an atom):


2aHC2H3O2 + aBa(OH)2 -> 2aH2O + aBa(C2H3O2)2
Now all you have to do is chose a number for a, in our case, we will chose 1 to be a (a = 1). Now just plug in 1 where you see "a" in the reaction.

2(1)HC2H3O2 + 1Ba(OH)2 -> 2(1)H2O + 1Ba(C2H3O2)2
and here you have it, a balanced equation:

2HC2H3O2 + Ba(OH)2 -> 2H2O + Ba(C2H3O2)2

This method is a more systematic way of balancing equations. It is good for more complicated equations that cannot be balanced by simply using trial and error method.

[AmitRazdan]

bravo i remember doing this back in high school