The balanced reaction for the combustion of pentane is:

When balanced, oxygen gas has a coefficient of eight.

To balance the equation, it is easiest to leave oxygen and hydrogen for last. This means we should start with carbon.

Now that carbon is balanced, we can look at hydrogen.

Finally, we can balance the oxygen.

The final reaction uses five carbon atoms, twelve hydrogen atoms, and sixteen oxygen atoms per side.

Potassium is a Group I element, so to get to a filled valence shell, it will lost one electron, yielding .

Arsenic is a Group 5 element, so it needs to gain three electrons to obtain a filled valence shell, yielding .

In order to balance out the charges, the resultant salt will be .

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:

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

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.

Iron (II) has a positive two charge: .

Phosphate has a negative three charge: .

The initial compound must be constructed to cancel these charges. The dissociation is: .

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.

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.