Limiting Reagents
WE BE COOKING…. OH CRAP I ONLY HAVE ½
CUP OF SUGAR!
Brownies again…
 Only have ½ cup of sugar…bummer….what to do?
½ cup butter
2 oz of chocolate
1 cp sugar
2 eggs
1 tsp vanilla
2/3 cp flour
½ tsp baking powder
¼ tsp salt
24 Brownies
Need to do some Stoiching to
figure out how much of other
ingredients if only have ½ cup
sugar….
Sugar is the limiting reagent
Everything else in excess.
Limiting Reagent
 Demo
 Solid zinc is added to 6.00 M HCl
 Zinc will all dissolve but….
 pH is still below 7…meaning still have H+ ions present
 Which reagent is limiting?
 Zinc
 Which reagent is excess?
 HCl
Stoichiometry with Limiting Reagent
 How to determine the limiting and excess reagent
 1. Get both reagents into the unit moles
 2. Divide those moles found by the coefficient of the reagent
 3. The smaller number is the limiting reagent (LR)
 4. The larger number is in excess
 5. All stoich problems will be determined by the limiting
reagent (LR)…therefore the given for DA will be the LR
Example with Limiting Reagent
7.24 moles of magnesium is added to 3.86 moles of oxygen
gas to make MgO. Which reactant is used up? Excess?
2Mg(s) + O2(g)  2MgO(s)
7.24 mol Mg
2
3.86 mol O2
1
= 3.62 mol of Mg
Mg is LR
= 3.86 mol of O2
O2 is excess
7.24 mol of Mg controls the reaction, therefore 7.24
mol will be your given for any DA
Example with Limiting Reagent
7.24 moles of magnesium is added to 3.86 moles of oxygen
gas to make MgO. How many grams of MgO is produced?
How many grams of oxygen is needed?
2Mg(s) + O2(g)  2MgO(s)
7.24 mol Mg
7.24 mol Mg
2 mol MgO
40.304 g MgO
2 mol Mg
1 mol MgO
1 mol O2
31.998 g O2
2 mol Mg
1 mol O2
= 292 g of MgO
= 116 g of O2
Remember…
 Must work with the Limiting Reagent in Stoich
 Must get both reagents into moles
 Divide each by their coefficient
 The smaller number is the LR…controls the reaction

This will be your given in DA
Problem
If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M
hydrochloric acid, HCl, solution, hydrogen gas and iron(II)
chloride are produced. Which reactant is limiting? How
many grams are in excess? How many grams of each
product is formed?
1.
Balance chemical reaction
Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq)
Problem
If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M
hydrochloric acid, HCl, solution, hydrogen gas and iron(II)
chloride are produced. Which reactant is limiting? How many
grams are in excess? How many grams of each product is
formed?
Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq)
2. Find moles of each reactant to find LR
7.56 g Fe
1 mol Fe
55.845 g
100. mL
1L
1000 mL
= 0.135 mol Fe
= 0.135 mol Fe
1
1.00 mol HCl = 0.100 mol HCl
1L
2
LR
= 0.0500 mol HCl
Problem
If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M
hydrochloric acid, HCl, solution, hydrogen gas and iron(II)
chloride are produced. Which reactant is limiting? How many
grams are in excess? How many grams of each product is
formed?
Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq)
0.100 mol HCl is LR
3. Find grams of excess (find grams you actually need)
0.100 mol HCl
1 mol Fe
55.845 g Fe
2 mol HCl
1 mol Fe
= 2.79 g of Fe
7.56 gram Fe – 2.79 gram Fe = 4.80 gram Fe excess
Problem
If 7.56 gram of iron metal are placed in 100. mL of a 1.00 M
hydrochloric acid, HCl, solution, hydrogen gas and iron(II)
chloride are produced. Which reactant is limiting? How many
grams are in excess? How many grams of each product is
formed?
Fe(s) + 2HCl(aq)  H2(g) + FeCl2(aq)
0.100 mol HCl is LR
3. Find grams of each product (H2 and FeCl2)
0.100 mol HCl
0.100 mol HCl
1 mol FeCl2
126.745 g FeCl2
2 mol HCl
1 mol FeCl2
1 mol H2
2.0158 g H2
2 mol HCl
1 mol H2
= 6.34 g of FeCl2
= 0.101 g of H2
Review Steps for Stoiching…
 Write and balancing chemical reaction
 Find limiting reagent
 Determine moles of each reactant
 Divide the moles be their coefficient
 The smaller number is LR
 Use the moles of the LR to convert to anything else
in the problem

Moleland!!
 Convert to moles of another compound/atom
 Convert to final unit (moles/grams/volume)
Percent Yield
Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)
How many grams of copper will be required to completely
mass
replace silver from 208 mL of 0.100Coefficients
M solution ofMolar
AgNO
3?
of Cu
208 mL
1L
1000 mL
0.100 mol AgNO3 1 mol Cu
1L
2 mol AgNO3
= 0.661 g of Cu
63.546 g Cu
1 mol Cu
Percent Yield
 100% is maximum
 Percent yield measures how well you did the lab
 Compares lab measurement (actual) over theoretical
amount (found using Stoich)
Actual amount (grams or moles) [Lab]
% Yield =
Theoretical amount (grams or moles)
[Stoich]
x 100
Percent Yield
If 12.5 grams of copper are reacted with an excess of chlorine
gas, then 25.4 grams of copper(II) chloride, CuCl2(s), are
obtained. Calculate theoretical amount and percent yield.
Cu(s) + Cl2(g)  CuCl2(s)
Find grams CuCl
2 mass
Molar
Coefficients
of Cu
12.5 g Cu
Molar mass
of CuCl2
1 mol
1 mol CuCl2
134.446 g CuCl2
63.546 g
1 mol Cu
1 mol CuCl2
25.4 g
% Yield =
26.4 g
x 100 = 96.4%
= 26.4 g of CuCl2