Combustion is the chemical reaction of a hydrocarbon with molecular oxygen, and it always produces carbon dioxide and water. Knowing the reactants and products, the unbalanced equation must be: 

We start by balancing the hydrogens. Since there are 10 on the left and only 2 on the right, we put a coefficient of 5 on water.

Similarly, we balance carbons by putting a 4 on the carbon dioxide.

To find the number of oxygens on the right, we multiply the 4 coefficient by the 2 subscript on O (which gets us 8 oxygens) and then add the 5 oxygens from the 5 water molecules to get a total of 13. The needed coefficient for  on the left would then have to be 13/2.

Because fractional coefficients are not allowed, we mutiply every coefficient by 2 to find our final reaction:

When we check the activity series, it is fairly easy to see that aluminum metal is more reactive than zinc metal. So, in this case, the two metals undergo a redox reaction, where the aqueous  is reduced to solid , and the solid  is oxidized to aqueous . These charges are the common oxidation states for zinc and aluminum and should be memorized.

Because  is the new species, it bonds with 3  ions. The unbalanced equation is:

We note that there are 2 chlorine atoms on the left and 3 chlorine atoms on the right. To balance, we use a 3 coefficient on the left and a 2 coefficient on the right. This gives a total of 6 chlorine atoms on eahc side.

However, now we have also increased the amounts of zinc and aluminum. We copy the necessary coefficients to balance those—2 for aluminum on the left, 3 for zinc on the right—and we are done:

Balancing reactions is best acheived by using a stepwise approach.  It is useful to work through each atom making sure it is present in a balanced fashion on both sides of the equation.  Starting with Fe, Fe₂O₃ is the only molecule with Fe present on the left side of the equation and FeCl₃ is the only molecule with Fe present on the right side of the equation.  Thus the molar ratio of Fe₂O₃ to FeCl₃ must be 1:2.  Moving on to O, the O on the left side of the equation exists as Fe₂O₃ and H₂O on the right.  The molar ratio of Fe₂O₃ to H₂O is therefore 1:3.  Cl exists in HCl on the left and FeCl₃ on the right. Thus the molar ratio of HCl to FeCl₃ must be 3:1. To ensure that these molar ratios are maintaned the balanced formula is then determined to be Fe₂O₃ + 6HCl -> 2FeCl₃ + 3H₂O

The first thing to do is to balance the Cl (3 on the left; one on the right) ← add 3 for NH₄Cl

Now, there are 3 N on the right and only one on the left; add a 3 to the NOH₅

Check to see that the Fe, H, and O balance, which they do

The reaction balances out to 2C₂H₆(s) + 7O₂(g) → 6H₂O(l) + 4CO₂(g). For every  2 moles of  C₂H₆ you can produce 6 moles of H₂O. Giving you a total production of 12 moles when you have 4 moles of C₂H₆.

When balancing equations, the goal is to make sure that the same atoms, in both type and amount, are on both the reactant and product side of the equation. A helpful approach is to write down the number of atoms already on both sides of the unbalanced equation. This way, you can predict which compounds need to be increased on which side in order to balance the equation. It also helps to balance oxygen and hydrogen last in the equation.

In this reaction, we can balance as follows.

Reactants: 1K, 1Mn, 1Cl, 4O, 1H

Products: 1K, 1Mn, 5Cl, 1O, 2H

So, we will need to increase H₂O and HCl. The final balanced equation is written below.

To balance an equation, we need to make sure there is the same amount of elements to the left of the arrow as there is to the right. We also need all the charges to balance out. We notice right away that there are three chlorine atoms on the left, but only one on the right.

(1Na, 1O, 1H, 1Fe, 3Cl : 1Na, 3O, 3H, 1Fe, 1Cl)

We can solve this by multiplying NaCl by three.

 (1Na, 1O, 1H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 3Cl)

This causes us to have an imbalance of sodium, which we can correct by manipulating NaOH.

 (3Na, 3O, 3H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 1Cl)

This is the final balanced equation. Note that it is usually easiest to manipulate oxygen and hydrogen last, since they are often involved in multiple molecules.

Begin by balancing the equation. There are many ways to do this, but one method that is particularly useful is to assume that you have 1 mole of your hydrocarbon (propane), and balance the equation from there. It may be necessary to manipulate an equation further if you end up with fractions, but all you will need to do is multiply by an integer if that is the case.

First, we will balance the carbons. There are three carbons in propane, so we will make sure there are three carbons on the right side of the arrow as well:

This is not complete. We will next balance the hydrogens. There are eight hydrogens on the left side of the equation, so:

The last step is to balance the oxygens on the left and right side of the equation

Our equation is balanced and all coefficients are integers. If we begin with one mole of propane, we will produce four moles of water.

First, we will balance the equation:

Since chlorine is in excess, we know that the limiting reagent is iron.

First, balance the carbon atoms in the reactants and products. Next, balance the hydrogen atoms. Third, balnace the oxygen atoms. Since stoichiometric coefficients are written as integers, double everything to remove the decimal/fraction.