Permalink Submitted by Boron on Tue, 2017-03-28 12:35

Hi several essential concepts. First a mole is an amount of something. Like a dozen eggs is an amount of something. But a mole is a very large number because it tells us how many of something very small there is in an amount that is 'macroscopic'. a mole of something has Avogadros number which is 6.02 x 10^23. So a dozen paperclips is 12 paperclips. A mole of carbon atoms is Avogadros number of carbon atoms. From the Formula Weight of something we know how many grams per mole of that 'something'. A sample of 100% carbon 12 (6 protons and 6 neutrons) is defined as 12 g/mole I believe. From the periodic table we get the FW of all of the different atoms and they are based on natural distributions of isotopes. This is probably on the inside cover of most chemistry books I am guessing; there is a periodic table with the weights of the atoms. But carbon here is not 12 I think it's 12.01 because in a sample of carbon atoms whether in elements (diamonds, graphite) or molecules (carbon dioxide, sugar, octane) there are other isotopes of carbon than carbon 12.

So with a formula weight for an atom or molecule you know how much grams in a mole. For example water's formula weight (H2O) is (1.008 x2) + (16.00) I believe.

So in your question you have 25 g Mg. Using what we have learned we should be able to find the moles of Mg that you use to react.

Ok now next concept is the balanced formula equation. This tells us what is happening in our reaction. For every 3 atoms of Mg we form 2 atoms of Fe. And since a mole is simply a large number of something (Avogodros number) we also know that for every 3 moles of Mg we form 2 moles of Fe.

Ok so a flow chart for a proposed strategy to solve the problem:

1) mass Mg (g) ----> amount Mg (moles) comment: use the formula weight to solve this.

2) amount Mg (moles) ----> amount Fe (moles) comment: use the ratio of Fe atoms (or moles) produced for each Mg atom (or mole) in the balanced equation.

Solution:

amount Mg (moles) = 25 g Mg x 1/24.305 (moles Mg/g Mg) = 1.0 moles (2 sig figs from 2 sig figs in 25 g)

amount Fe (moles) = 1.0 moles Mg x (2/3) (moles Fe/moles Mg) = .67 moles Fe

comment: the 2 moles Fe / 3 moles Mg comes from the balanced equation coefficients in front of Fe and Mg

comment: the 2/3 relationship is an exact relationship so it is not ever (I believe) the limiting factor on sig figs.

## Hi several essential concepts

Hi several essential concepts. First a mole is an amount of something. Like a dozen eggs is an amount of something. But a mole is a very large number because it tells us how many of something very small there is in an amount that is 'macroscopic'. a mole of something has Avogadros number which is 6.02 x 10^23. So a dozen paperclips is 12 paperclips. A mole of carbon atoms is Avogadros number of carbon atoms. From the Formula Weight of something we know how many grams per mole of that 'something'. A sample of 100% carbon 12 (6 protons and 6 neutrons) is defined as 12 g/mole I believe. From the periodic table we get the FW of all of the different atoms and they are based on natural distributions of isotopes. This is probably on the inside cover of most chemistry books I am guessing; there is a periodic table with the weights of the atoms. But carbon here is not 12 I think it's 12.01 because in a sample of carbon atoms whether in elements (diamonds, graphite) or molecules (carbon dioxide, sugar, octane) there are other isotopes of carbon than carbon 12.

So with a formula weight for an atom or molecule you know how much grams in a mole. For example water's formula weight (H2O) is (1.008 x2) + (16.00) I believe.

So in your question you have 25 g Mg. Using what we have learned we should be able to find the moles of Mg that you use to react.

Ok now next concept is the balanced formula equation. This tells us what is happening in our reaction. For every 3 atoms of Mg we form 2 atoms of Fe. And since a mole is simply a large number of something (Avogodros number) we also know that for every 3 moles of Mg we form 2 moles of Fe.

Ok so a flow chart for a proposed strategy to solve the problem:

1) mass Mg (g) ----> amount Mg (moles) comment: use the formula weight to solve this.

2) amount Mg (moles) ----> amount Fe (moles) comment: use the ratio of Fe atoms (or moles) produced for each Mg atom (or mole) in the balanced equation.

Solution:

amount Mg (moles) = 25 g Mg x 1/24.305 (moles Mg/g Mg) = 1.0 moles (2 sig figs from 2 sig figs in 25 g)

amount Fe (moles) = 1.0 moles Mg x (2/3) (moles Fe/moles Mg) = .67 moles Fe

comment: the 2 moles Fe / 3 moles Mg comes from the balanced equation coefficients in front of Fe and Mg

comment: the 2/3 relationship is an exact relationship so it is not ever (I believe) the limiting factor on sig figs.