STOICHIOMETRY
Stoichiometry is the relation between the masses of the reactants and products in chemical reaction. In
stoichiometry Chemical reactions must be balanced, or in other words, must have the same number of
various atoms in the products as in the reactants. Also, the amounts obtainable from a balance equation
are called stoichiometric amounts.
We already know what is stoichiometry and so we’re going to solve a problem.
Aluminum reacts with Chlorine gas to form Aluminum Chloride. (a) If 35g of Aluminum reacts with
excess Chlorine how many grams of Aluminum Chloride will form? (b) How many grams of Chlorine will
react completely with 42.8g of Aluminum.
To solve this problem, we have to:
Translate the problem into chemical equation.
Balance the chemical equation.
Interpret the balanced chemical equation, what information can be obtained from the balanced
chemical equation?
Al + Cl₂ AlCl₃
2Al + 3Cl₂ 2AlCl₃
a.
Given: 35 g Al
26.98 g Al
Al + 3Cl = 26.98 + 3(35.45) = 133.33
1 mol Al 2mol AlCl ₃ 133.33 g AlCl ₃
35 g Al x x x = 172.96 g AlCl₃
26.98 g Al 2 mol Al 1 mol AlCl ₃
b.
Given: 42.8 g Al
26.98 g Al
mm= Cl₂
=35.45 x 2
= 70.9 g Cl₂
1 mol Al 3 mol Cl ₂ 70.9 g Cl ₂
42.8 g Alx x x = 169.84 g Cl₂
26.9 8 g Al 2 mol Al 1 mol Cl ₂
LIMITING AND EXCESS REACTANT
Limiting Reagent, the limiting reagent in a chemical reaction is the reactant that will be consumed
completely. Once there is no more of that reactant, the reaction cannot proceed. Therefor it limits the
reaction from continuing. Excess Reagent The excess reagent is the reactant that could keep reacting if
the other had not been consumed Limiting reactant and theoretical yield
Problem: If 14.32g of N₂ reacts of 4.21g of H₂ to produce NH₃, what is the limiting reactant?
To solve the problem, we must:
Step 1: Convert grams of each reactant to grams of product
Step 2: Find the limiting reactant
To find the limiting reactant we must divide the two given chemical molar masses in how many molar
masses they have. After we solve the lowest answer will become the limiting reactant.
Solution:
N₂ + 3H₂ 2NH₃
mm of N₂= 2(14.01) = 28.02 g/mol
mm of H₂= 2(1.008) = 2.016 g/mol
14.32 g
nN₂ = = 0.511 mol
28.01 g /mol
4.21 g
nH₂ = = 2.088 mol
2.016 g /mol
0.511
For N₂= = 0.511 Coefficient of N₂
1
2.088
For H₂= = 0.696 Coefficient of H₂
3
Limiting Reactant = N₂
Excess Reactant = H₂
The Theoretical and Percent yield
As we just learned, the theoretical yield is the maximum amount of product that can be formed in a
chemical reaction based on the amount of limiting reactant. In practice, however, the actual yield of
product—the amount of product that is actually obtained—is almost always lower than the theoretical
yield. This can be due to a number of factors, including side reactions (secondary reactions that form
undesired products) or purification steps that lower the amount of product isolated after the reaction.
Actual Yield
Percent Yield = x 100% =
Theoretical Yield
To solve the percent yield, we must first obtain the actual yield. We can easily obtain the actual yield
because it is provided in the problem. After we have solved the problem and obtained the theoretical
yield, we can divide the actual yield by the theoretical yield. After we have the answer, multiply it by 100
to get the percent yield.
Problem: In the reaction vessel, 18.0g of N₂ and 8.00g of H₂ were allowed to react to form ammonia. If
15.0g NH₃ was recovered after the reaction, find (a) The Theoretical Yield (b) Percent Yield.
N₂ + 3H₂ 2NH₃
A.
18.0 g
nN₂ = = 0.643
28 g /mol
8.00 g
nH₂ = = 3.960
2.02 g /mol
0.643
For N₂ = = 0.643 (lesser)
1
3.960
For H₂ = = 1.32
3
Limiting Reactant= N₂
1mol of N ₂ 2mol of NH ₃ 17.03 g of NH ₃
Theoretical Yield= 18.0 g of N₂ x x x = 21.90 g of NH₃
28 g of N ₂ 1 mol of N ₂ 1 mol of NH ₃
B.
Actual Yield
% Yield = x 100%
Theoretical Yield
15.0 g
= X 100%
21.90 g
% Yield = 68.49%
