Balancing equations:

there is 1 NaOH and 1 H₂SO₄ in the reactants; and 1 H₂O and 1 Na₂SO₄ in the products

steps:

1) balance the Na first; get 2 NaOH

2) balance H next; get 1 H₂SO4, 2 H₂O

3) check to make sure that O and S balance

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.

\(\displaystyle 2KMnO_{4} + 16HCl \rightarrow 2KCl + 2MnCl_{2} + 8H_{2}O + 5Cl_{2}\)

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.

\(\displaystyle NaOH + FeCl_{3} \rightarrow NaCl + Fe(OH)_3\) (1Na, 1O, 1H, 1Fe, 3Cl : 1Na, 3O, 3H, 1Fe, 1Cl)

We can solve this by multiplying NaCl by three.

\(\displaystyle NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3\) (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.

\(\displaystyle 3NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3\) (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.

Calcium is in the second group of the periodic table, and is therefore going to have a \(\displaystyle \small +2\) oxidation number. Hydroxide ions have a \(\displaystyle \small -1\) charge. Calcium hydroxide will have the formula \(\displaystyle \small Ca(OH)_2\).

Chloride ions have a \(\displaystyle \small -1\) charge and hydrogen ions have a \(\displaystyle \small +1\) charge. The formula for hydrochloric acid is \(\displaystyle \small HCl\).

On the products side, water has the formula \(\displaystyle \small H_2O\) and calcium chloride has the formula \(\displaystyle \small CaCl_2\).

Now that we know all of the formulas, we can write our reaction:

\(\displaystyle Ca(OH)_2+HCl\rightarrow H_2O+CaCl_2\)

In order to balance the chloride atoms, we need to add coefficients.

\(\displaystyle Ca(OH)_2+2HCl\rightarrow 2H_2O+CaCl_2\)

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:

\(\displaystyle C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O\)

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

\(\displaystyle C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O\)

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

\(\displaystyle C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\)

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

\(\displaystyle Fe+Cl_2\rightarrow FeCl_3\)

First, we will balance the equation:

\(\displaystyle 2Fe+3Cl_2\rightarrow 2FeCl_3\)

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

\(\displaystyle 36.0\hspace{1 mm}g\hspace{1 mm}FeCl_3\times\frac{1\hspace{1 mm}mole\hspace{1 mm}FeCl_3}{162.2\hspace{1 mm}g\hspace{1 mm}FeCl_3}\times\frac{2\hspace{1 mm}moles\hspace{1 mm}Fe}{2\hspace{1 mm}moles\hspace{1 mm}FeCl_3}\times\frac{55.845\hspace{1 mm}g\hspace{1 mm}Fe}{1\hspace{1 mm}mole\hspace{1 mm}Fe}=12.4\hspace{1 mm}g\hspace{1 mm}Fe\)

In the balanced version of the equation:

\(\displaystyle S + 6HNO_{3} \rightleftharpoons H_{2}SO_{4} + 6NO_{2} + 2H_{2}O\)

Nitrogen dioxide's coefficient is 6 in order to balance the 6 moles of nitrogen provided by the 6 moles of nitric acid on the equation's left side.