Stoichiometry - Chemistry - page 217

2MgO + O 2 2MgO

2.00 g excess x g

mass of sample
molar mass

2.00 g
24.3 g/mole

# of moles of magnesium = = = 0.0823 moles of Mg

So, we start with 0.0823 moles of magnesium and an excess of oxygen,
and we want to find out the mass of the magnesium oxide produced. The
next step is to compare the molar ratio shown by the coefficients in the
balanced chemical reaction. Do you notice that both the magnesium and
the magnesium oxide are preceded by coefficients of 2? This means that
for every 2 moles of magnesium that you burn, you will produce 2 moles of
magnesium oxide, provided you have an excess of oxygen. In other words,
the molar ratio of Mg to MgO is 2:2 or 1:1. This tells us that if we put
0.0823 moles of Mg into the reaction, we will get 0.0823 moles of MgO out
of the reaction.
Now that we know how many moles of MgO we will produce, we must
convert that to grams by multiplying the number of moles of MgO (0.0823
moles) by the molar mass of MgO (24.3 g/mole + 16.0 g/mole = 40.3 g/mole),
which we get by adding the molar mass of magnesium and oxygen.
2MgO + O 2 2MgO
2.00 g excess xg

mass of sample
molar mass

2.00 g
24.3 g/mole

# of moles of magnesium = = = 0.0823 moles of Mg

Mass of MgO = # of moles × molar mass = 0.0823 moles × 40.3 g/mole = 3.32 g

So, 3.32 g of MgO are produced when 2.00 g of Mg react with an
excess of oxygen.

In summary, the steps for solving this type of problem are:
1. Begin with a balanced chemical reaction.

2. Carefully label the equation with the information from the

word problem.

3. Convert the given quantity to number of moles.

4. Use the molar ratio shown by the coefficients to determine

the number of moles of the unknown.

5. Convert this number of moles to the quantity (grams, in the

case of this example) the question asks for.