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. 

One molecule of , balances with  molecules and  molecules. On the right side of the equation, there are a total of 18 oxygen atoms, which equates to  molecules on the left side of the equation.

As when balancing the vast majority of chemical reaction, leave oxygen as the last element to balance. Start by balancing the hydrogen atoms - this requires a 4 in front of water on the products side. Next, balance the three carbon atoms on the reactant side by adding a coefficient of 3 to the carbon dioxide in the products. Finally balance the oxygen atoms by placing a 5 in front of the oxygen gas on the reactant side. Double check to make sure there is a 1:1 ratio of each element and that all coefficients are whole numbers. Thus the balanced chemical reaction is: 

Balance aluminum by adding a coefficient of 2 to the left side:

Balance iron by adding a coefficient of 2 to the right side:

The equation is balanced.