1.1 What is Chemistry?

12.3. Limiting Reactant and Percent Yield http://www.ck12.org

TABLE12.3:Magnesium Table Part I

Mg O 2 MgO

Molar Mass 24.31 g/mol 32.00 g/mol 40.31 g/mol

Initial Mass 1.00 g excess 0 g

Initial Moles excess 0 mol

Change in Moles -2x -x +2x

Final Moles 0 mol excess

Final Mass 0 g excess

Now, fill in the remaining values by performing calculations. We can use molar masses to convert between grams
and moles for any known amount.

1 .00 g Mg( 24 1 mol Mg.31 g Mg) = 0 .0411 mol Mg

Once we know the difference between the initial and final moles for any of the reaction components, we can calculate
the value of x. Over the course of the reaction, 0.0411 moles of Mg are used up, so

− 2 x=− 0 .0411 mol

x= 0 .0206 mol

Using this information, we can fill in more of the table:

TABLE12.4:Magnesium Table Part II

Mg O 2 MgO

Molar Mass 24.31 g/mol 32.00 g/mol 40.31 g/mol

Initial Mass 1.00 g excess 0 g

Initial Moles 0.0411 mol excess 0 mol

Change in Moles -0.0411 mol -0.0206 mol +0.0411 mol

Final Moles 0 mol excess

Final Mass 0 g excess

Using our value for x, we can now perform simple addition to determine the final moles of MgO. This can then be
converted to grams using the molar mass.

TABLE12.5:Magnesium Table Part III

Mg O 2 MgO

Molar Mass 24.31 g/mol 32.00 g/mol 40.31 g/mol

Initial Mass 1.00 g excess 0 g

Initial Moles 0.0411 mol excess 0 mol

Change in Moles -0.0411 mol -0.0206 mol +0.0411 mol

Final Moles 0 mol excess 0.0411 mol

Final Mass 0 g excess 1.66 g

Now that the table is complete, we can answer the original questions. We can see directly from the table that 1.66
grams of MgO could be produced from this reaction. We do not yet know the mass of oxygen that is used up, but
we do know that 0.0206 moles are consumed. Converting this to grams using the molar mass of O 2 gives us the