When performing a mass to mass stoichiometry problem, explain why masses must be converted to moles rather than directly using the mass of reactants to predict the mass of products. Consider the synthesis reaction of Mg and O2 below. If equal masses of these reactants are allowed to react, explain how you would know which reactant is the limiting reactant and which reactant is the excess reactant. (Lesson 8)

2 Mg + O2 → 2 MgO

A: When performing a mass to mass stoichiometry problem, we must convert mass to moles because each compound has a different molecular wieght. By each compound having a different mass we need to convert the mass to moles which will allow for the necessary calculations.

If equal masses of these reactants were allowed to react Mg is the limiting reactant, while O2 is the excess reactant. First the limiting reactant is the element used up in a chemical reaction (Mg) and an excess reactant is the reactant that will be left when the limiting reactants (O2) has been used up. So, if equal masses of these reactants were able to react, meaning 1 gram of Mg (Magnesium) and 1 gram of O (oxygen), Mg would be the limiting reactant because Mg weights more than O. Therefore there would be less atoms in the 1 gram of Mg to react with 1 gram of oxygen.

2 Mg + O2 → 2 MgO

        1 mol O                 2 MgO       40.3044 gMgO

10 x -----------------  x    ----------- x   ------------------ = 25.19O2=25.2gO2

       32gO (16*2)           1 Mol           1 Mol O

        2 mol Mg          2 MgO       40.3044 gMgO

10 x -------------  x    ----------- x   ------------------ = 16.7935= 16.8gMg

       48gMg               2 Mol           1 Mol O2

Q: Explain the process used to balance equations and then give the coefficients that will balance the equation below. (Lesson 7) AlCl3 + K2SO4 → Al2(SO4)3 + KCl.

A: In order to have a balance equation, the same number of atoms in the product and the reactant, we must first check each side of the equation (reactant side, AlCl3 + K2SO4 and the product side, Al2(SO4)3 + KCl).

Once this has been completed, we can determine the equation above is not balanced due to the product side not being equal to not equal to the reactant side. The reactant side shows Cl3 (Chlorine) and K2 (Potassium) while the product side shows only KCl only. Next you notice the product side show Al2 and (SO4)3 while the reactant side shows Al and SO4.

The next we will begin to balance the above equation by counting the atoms of each element for the reactant/product side.

AlCl3 + K2SO4 → Al2(SO4)3 + KCl

Reactant  Product

1 Al         2 Al Not Balanced

3 Cl         1 Cl Not

2 K          1 K Not

4 SO      12 SO Not

After listing the elements we need to add coefficients to balance each element without changing any of the subscripts.

AlCl3 + K2SO4 → Al2(SO4)3 + KCl

We need to add a coefficient of 2 and 3 to the reactant side so it reads 2AlCl3 + 3K2SO4

Next we can add a coefficient of 6 on the product side so it reads Al2(SO4)3 + 6KCl making this equation balanced.

2AlCl3 + 3K2SO4 → Al2(SO4)3+6KCl

To check if the equation is balanced list the total atoms

Reactant  Product

2 Al        2 Al Balanced

6 Cl        6 Cl Balanced

6 K         6 K  Balanced

12 SO    12 SO   Balanced