When balancing a chemical equation, it means that we are adjusting the coefficients in front of every compound so that the number of atoms for every element will match how many will be on the other side of the arrow.

Initially, we must take "inventory" of how many atoms of each element we have on both sides of the arrow.

On the left side:

1 iron, 2 chlorine

On the right side:

1 iron, 3 chlorine

While the iron atoms are balanced, the chlorine atoms are not. Now we must think, what is the least common multiple of 2 and 3? We can quickly realize it is 6. So we must add a coefficient in front of every term that includes a chlorine so we may make it 6 chlorine atoms on both sides of the arrow. (The coefficient will be multiplied by the subscript attached to the chlorine atom.)

While the chlorine atoms are balanced now, we have lost the balance for the iron atoms. This can be quickly resolved. On the left, we still only have one iron atom. On the right, it has changed to 2 iron atoms. This means that in order to balance the number of iron atoms, we just need to add a 2 coefficient in front of the iron on the left side of the arrow. 

Now let's do one last tally to make sure we've balanced everything.

On the left:

2 iron, 6 chlorine

On the right:

2 iron, 6 chlorine

The equation has been balanced. 

When balancing a chemical equation, it means that we are adjusting the coefficients in front of every compound so that the number of atoms for every element will match how many will be on the other side of the arrow.

Initially, we must take "inventory" of how many atoms of each element we have on both sides of the arrow.

On the left side:

1 potassium, 5 oxygen, 4 hydrogen, 1 P

On the right side:

3 potassium, 5 oxygen, 2 hydrogen, 1 P

Since there are quite a few components in this example, all of which need balancing, we just need to choose one element to start with and continue from there. Let's look at potassium. Since there is only one potassium atom on the left side, and 3 on the right side, let's think about what the lowest common multiple between 1 and 3 is - we can quickly decide that it is 3. This means we may place a 3 coefficient in front of the  on the left side of the arrow. (The coefficient is multiplied by the subscript of the atom - in this case, 3 multiplies by 1.)

Adding the 3 coefficient in front of  changes how many oxygen and hydrogen atoms we have on the left side to 7 oxygen atoms and  6 hydrogen atoms. Comparing to the right side, which has 5 oxygen atoms and 2 hydrogen atoms. Since balancing the oxygen atoms on both sides is slightly more complicated because oxygen appears in two compounds, we will not manipulate their numbers but let their numbers become balanced through balancing the atoms around them. So let's turn our attention to the hydrogen atoms. If we have 6 hydrogen atoms on the left side, and 2 hydrogen atoms on the right (in one compound), we can come to terms with 6 being the lowest common multiple for 6 and 2. This means we need to place a 3 coefficient in front of  on the right side. 

This brings our total of oxygen atoms on the right side to 7 as well. So we have indirectly balanced the oxygen atoms. Since the only element left to balance is phosphorus, we realize that phosphorus on both sides is already balanced. We have balanced our equation. 

Now let's do one last tally to make sure we've balanced everything.

On the left:

3 potassium, 7 oxygen, 6 hydrogen, 1 phosphorus

On the right:

3 potassium, 7 oxygen, 6 hydrogen, 1 phosphorus

The equation has been balanced.

Since we are given the mass of nitrogen gas that will be used, we will need to convert the amount into moles. We can then compare the molar ratios in the balanced reaction, and multiply the moles of ammonia by its molar mass. Using dimensional analysis, we can use a calculation that will allow us to end with "grams of ammonia" as the solved unit.

The formula for glucose is _.

First, we must find the molecular weight of the entire molecule of glucose. We do this individually for each element. Using oxygen as an example, we multiply the atomic mass (16) by the number of atoms per glucose molecule:

We do similar calculations for carbon and hydrogen to find our total molecular weight for glucose.

Sum these values to find the molecular weight.

Next, to find percent composition by mass for oxygen, we are being asked to find how much of one molecule of glucose's weight comes from oxygen. To do this we divide the mass of oxygen by the molecular weight:

Thus 53.3% of the mass of glucose comes from oxygen alone.

First, convert the number of atoms to moles using Avogadro's number:

Next, convert moles of iron to grams of iron using the molar mass:

To find the number of moles in a sample, use the periodic table to find the molar mass. Since the atomic weights of  and  are  and , respectively, add the two atomic masses to determine the molar mass of

Next, use dimensional analysis to find the number of moles in the given mass of the sample.

Use the periodic table to compute the molecular mass of .

Next, use dimensional analysis to convert grams to moles.

Use the correct number of significant figures.

First convert the 7.2mg of silicon into grams:

Then convert grams of silicon into moles of silicon:

Finally, convert moles of silicon into atoms using Avogadro's number:

Since 7.2mg of silicon is much less than 1 mole (which we know weighs 28.09g) our answer makes sense as it is much less than Avogadro's number.

The answer is not present, but  equates to .

Use Avogadro's number to convert from atoms to moles:

Convert from moles of  to grams of :