How many oxygens are indicated: 18

Balance these equations:

Equation 1) Zn + HCl ---> ZnCl₂ + H₂

Upon examining this equation, you see that there is already one Zn on each side of the equation. We will attempt to leave it alone, if at all possible, since it is already balanced.

One the right side, we see two chlorines and two hydrogens, with only one of each on the left. Putting a two in front of the HCl doubles the number of chlorine and hydrogen on the left side.

This now leaves us with two chlorines and two hydrogens on each side of the arrow, making them both balanced.

Since the zinc was already balanced, the entire equation is now balanced.

Equation 2) KClO₃ ---> KCl + O₂

Start by noticing the the K and the Cl are ALREADY balanced in the skeleton equation. However, the oxygen is out of balance with three on the left and two on the right.

It is important to emphasize that the oxygen on the left will increase only in steps of three, while the oxygen on the right will increase only in steps of two. The question to ask yourself is "What is the least common multiple between 2 and 3?" The answer of course is six. We need six oxygens on each side of the equation. We use a two on the left side since 2 x 3 = 6 and we use a three on the right side since 3 x 2 = 6.

This causes the K and the Cl to become unbalanced, but putting a two in front of the KCl on the right side fixes that.

This problem is interesting because you focused on the oxygens first. Normally, oxygen is the last (or next-to-last) element to be balanced.

Equation 3) S₈ + F₂ ---> SF₆

An eight in front of the SF₆ will balance the sulfurs.

This now gives us 48 fluorines on the right-hand side, since 8 x 6 = 48. Use a 24 in front of F₂ since 24 x 2 also equals 48.

Equation 4) Fe + O₂ ---> Fe₂O₃

In the unbalanced equation, there was only one Fe on the left and two on the right. Putting a two in front of the Fe on the left brings the irons into balance.

The situation balancing the oxygen is quite common. You saw it in a previous example. This time, I'll try to lay it out in steps.

1. The oxygen on the left ONLY comes in twos, while the right-hand side oxygen comes only in threes.
2. We have to get an equal number of oxygens. (Remember, we can only adjust the value of the coefficient. We cannot change the subscript.)
3. The least common multiple between two and three is six. This means we will need six oxygens on each side of the equation.
4. To get this, we put a three in front of the O₂ since 3 x 2 = 6 and we put a two in front of the Fe₂O₃ since 2 x 3 = 6.

The Fe was balanced, but has become unbalanced as a consequence of our work with the oxygen. Putting a four in front of the Fe on the left solves this.

Equation 5) C₂H₆ + O₂ ---> CO₂ + H₂O First, balance the carbons with a two in front of the CO₂. Then balance the hydrogens with a three in front of the H₂O. This leaves the following equation:

Only the oxygens remain to be balanced, but there is a problem. On the right side of the equation, there are seven oxygen atoms, BUT oxygen only comes in a group of two atoms on the left side. Another way to say it - with O₂ it is impossible to generate an ODD number of oxygen atoms.

However, that is true only if you were using whole number coefficients. It is allowable to use FRACTIONAL coefficients in the balancing process. That means I can use seven-halves as a coefficient to balance this equation, like this:

Generally, the fractional coefficient is not retained in the final answer. Multiplying the coefficients through by two gets rid of the fraction. here is the final answer:

Also, improper fractions like 7/2 should be used rather than a mixed number like 3 1/2. Return