CH 115 Fall 2014
Worksheet 25
1. What are some signs that will indicate that you are working a problem with a
limiting reagent?
You are given two reactants and each one of them is present in significant
amounts. (E.g. 100 g of X reacts with 25 g of Y). Neither of these reactants is
labeled as being in excess.
2. What are the steps to working a limiting reagent problem? How do figure out which
reagent is the limiting reagent? How do figure out how much excess reagent is
remaining?
1. Make sure you have a balanced chemical equation.
2. Convert each reactant value to grams of product you are looking for.
3. Whichever reactant produced the least amount of product is your limiting
reagent.
4. The lesser value of product formed is the amount of product that
originated from the reaction.
5. To figure out how much excess reagent is remaining, first convert either
the grams of limiting reactant you started out with or the amount of
product formed to excess reagent – this will tell you how much of your
excess was used during the reaction. Then, subtract this value from the
amount of excess you started out with (given in problem).
3. A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction:
Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
How much MgCl2 is produced in this reaction? How much of the excess reactant is
left over when this reaction has run to completion?
This is a limiting reactant problem – we are given two reactant values to start
out with.
Convert each one of your reactants to grams of product (MgCl2).
1. 50.6 g Mg(OH)2 * (1 mol/58.31 g Mg(OH)2) * (1 mol MgCl2/1 mol Mg(OH)2) *
(95.21 g MgCl2/ 1 mol) = 82.62 g MgCl2
2. 45.0 g HCl * (1 mol HCl / 36.45 g HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.21 g
MgCl2 / 1 mol) = 58.77 g MgCl2
The limiting reagent is HCl and using it all up produces 58.77 g MgCl2.
45.0 g HCl * (1 mol HCl / 36.45 g HCl) * (1 mol Mg(OH)2 / 2 mol HCl) * (58.31 g
Mg(OH)2 / 1 mol) = 35.99 g Mg(OH)2 used
Excess remaining = 50.6 g – 35.99 g = 14.61 g Mg(OH)2
4. If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen, which is the
limiting reagent and how much excess reagent is leftover when this reaction is over?
First, we need a balanced equation. Combustion reaction, so CO2 and H2O are
produced.
C2H4 + 3 O2  2 CO2 + 2 H2O
4.95 g C2H4 * (1 mol C2H4 / 28 g C2H4) * (2 mol CO2 / 1 mol C2H4) * (44 g CO2 / 1
mol CO2) = 15.56 g CO2
3.25 g O2 * (1 mol O2 / 32 g O2) * (2 mol CO2 / 3 mol O2) * (44 g CO2 / 1 mol CO2)
= 2.98 g CO2
CH 115 Fall 2014
Worksheet 25
O2 is the limiting reagent (produced less product).
3.25 g O2 * (1 mol O2 / 32 g O2) * (1 mol C2H4 / 3 mol O2) * (28 g C2H4 / 1 mol
C2H4) = 0.948 g C2H4 excess reagent used
Excess remaining = 4.95 g – 0.948 g = 4.00 g
5. What is the formula for calculating percent yield?
% yield = actual yield (given in problem, determined experimentally)/
theoretical yield (calculated using stoichiometric techniques) x 100
6. If 15 grams of copper (II) chloride react with 20 grams of sodium nitrate by the
following equation and 11.3 grams of sodium chloride are produced, what is the
percent yield of sodium chloride for this reaction?
CuCl2 + 2 NaNO3  Cu(NO3)2 + 2 NaCl
Balance equation first!
Find theoretical yield using limiting reagent method.
15 g CuCl2 * (1 mol CuCl2 / 134.45 g CuCl2) * (2 mol NaCl / 1 mol CuCl2) * (58.45
g NaCl / 1 mol NaCl) = 13.04 g NaCl
20 g NaNO3 * (1 mol NaNO3 / 85 g NaNO3) * (2 mol NaCl / 2 mol NaNO3) * (58.45
g NaCl / 1 mol NaCl) = 13.75 g NaCl
Theoretical yield = 13.04 g
Percent yield = actual/theoretical * 100 = 11.3 g/13.04 g * 100 = 82.2%