Question

How does these chemical equation balance, step by step? Why do they balance this way?

I already know how they balnace I just don't understand how they balance.

why does the O(2*) balance to 25? 2 C(8*)H(18*) + 25 0(2) yields 16 CO(2*) + 18 H(2*)O

These ones have me stumped as well

1 C(3)H(8) + 5 O(2) = 3 CO(2) + 4 H(2)O

Why does this equation balance this way?

1 Pb(OH)(2) + 2 HCl = 2 H(2)O + 1 PbCl(2)

Pb(OH)(2) + HCl + H(2)O + PbCl(2)

As I look at this equation, I am immediatly stumped because of the (OH). It kind of looks like it affects the whole equation, but I don't know why or know.

Please explain to me, in depth, why and how these equations balance this way

Answer 1

Here's how you approach a combustion reaction problem.

C(3)H(8) + O(2) = CO(2) + H(2)O

Determine the molecule with the most atoms. In this case it is the C(3)H(8). Put a coefficient of "1" in front of it. This will be our base molecule which we will use to balance all the atoms.

C(3)H(8) has 3 carbons while CO has only 1 carbon; so, we need to place a coefficient of 3 in front of the CO

1C(3)H(8) + O(2) = 3CO(2) + H(2)O. Carbons are now balanced. Next we use our base molecule C(3)H(8) to determine the number of H atoms it contains. It has 8. On the right side of the equation we notice the only H atoms are in H(2)O, and there are 2 H atoms there.

We need to have 8 in order to balance the H atoms, so we place a coeffidient of 4 in front of the H(2)O. We need four (4) H(2)O molecules to get 8 H atoms.

1C(3)H(8) + O(2) = 3CO(2) + 4H(2)O. Now we have 3 carbons on the left side and 3 carbons on the right side of the equation; we also have 8 hydrogens on the left and 8 (4 x 2) hydrogens on the right.

That's all we can do with our base molecule, but we still need to balance the oxygen (O) atoms. Currently, we have 2 O atoms on the left and a total of 10 O atoms on the right (3x2 = 6 O atoms from the 3 CO(2) molecules and 4 O atoms from the 4 H(2)O molecules). We need to find a coefficient for the O(2) that will yield 10 O atoms to match the 10 O atoms on the right side of the equation. That's easy. Use a coefficient of 5 in front of the O(2).

5 x 2 = 10

1C(3)H(8) + 5 O(2) = 3CO(2) + 4H(2)O

Left side Right side

C= 2 x 3 = 6 C = 6 x 1 = 6

H= 2 x 8 = 16 H = 8 x 2 = 16

O = 5 x 2 = 10 O = (3 x 2) + (4 x 1) = 10

For your first problem, C(8)H(18) + O(2) = CO + H(2)O, it works in a similar way:

Choose C(8)H(18), place a coefficient of "1" in front of it.

1C(8)H(18) + O(2) = CO(2) + H(2)O; balance the C atoms - we need 8 on both sides

1C(8)H(18) + O(2) = 8CO(2) + H(2)O; balance the H atoms - we need 18 on both sides

1C(8)H(18) + O(2) = 8CO(2) + 9H(2)O; add up the oxygens on the left and compare with the number on the right side of the equation. We have 2 O atoms on the left and a total of 25 O atoms on the right [(8 x 2 = 16 from the CO molecules) + (9 x 1 = 9 from the H(2)O molecules)]. If we have 25 O atoms on the right, we need 25 on the left.

How do we do this? We use a coefficient of 25/2 on the left since 25/2 x 2 = 25.

1C(8)H(18) + 25/2 O(2) = 8CO(2) + 9H(2)O

We never leave fractions as coefficients, so we multiply the entire equation by 2

2C(8)H(18) + 25 O(2) = 16CO + 18H(2)O

For the: Pb(OH)(2) + HCl = H(2)O + PbCl(2), rewrite water [H(2)O] as H-OH

Pb(OH)(2) + HCl = H-OH + PbCl(2). Now balance the equation:

1Pb(OH)(2) + HCl = H-OH + PbCl(2); 1 Pb on the left, 1 Pb on the right

1Pb(OH)(2) + HCl = 2H-OH + PbCl(2); 1x2 =2 OH's on the left and 2x1=2 OH's on the right. Next, balance the Cl atoms.

1Pb(OH)(2) + 2HCl = 2H-OH + PbCl(2); Now we have 2 Cl atoms on the left, 2 Cl atoms on the right; 2 H atoms on the left, 2H atoms on the right; 2 OH groups on the left and 2 OH groups on the right; 1 Pb on the left, 1 Pb on the right.

We can now rewite the balanced equation with H-OH as H(2)O

Pb(OH)(2) + 2HCl = 2H(2)O + PbCl(2);