•Whenever two or more reactants are given in a chemical reaction, the limiting reactant must be identified before you can calculate the maximum amount of product that can be produced.
•A limiting reactant is completely used up in the chemical reaction.
•The other reactants are known as excess reactants and are not completely used up.
•6.00g of magnesium reacts with 2.00g of oxygen to form magnesium oxide.
•Determine the limiting reactant, the excess reactant (how many mole in excess) & mass of magnesium oxide formed.
Step 1: Write a balanced equation and identify known and unknown.
2Mg (s) + O2 (g) → 2MgO(s)
known known unknown
6.00 g 2.00 g
Step 2: Calculate n (known).
n(Mg) = n / M = 6.00 / 24.3 = 0.247 mol
n(O2) = n / M = 2.00 / 32.0 = 0.0625 mol
Compare mole ratio from equation:
Mg : O2
2 : 1
0.247 : 0.1235 (required)
Compare mole of O2 required to actual.
•Compare mole of O2 required to actual.
Required : Actual
0.1235 : 0.0625
•Because the actual mole is less than the required mole, there is not enough O2.
Therefore O2 is the limiting reactant.
•Therefore Mg is the excess reactant.
•To calculate the mole of Mg in excess use equation to compare mole ratios.
Mg : O2
2 : 1
0.125 : 0.0625
•0.125 mole of Mg reacts with all of the O2
•n (Mg) in excess
= n (Mg) initial – n (Mg) reacted
= 0.247 – 0.125
= 0.122 mol
Step 3: From equation find ratio of n(unknown) to n(known).
n(MgO) = 2/1 x n(O2)
= 2/1 x 0.0625
= 0.125 mol
Step 4: Answer Question
Find mass of magnesium oxide.
= n x M
= 0.125 x (24.3 + 16.0)
= 0.125 x 40.3
= 5.04 g