Transcript Limiting & Excess Reactants

Limiting Reactants – Method 1

1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
◦ limiting reactant
◦ amount of product
Limiting Reactants
How much product would be made if we begin
with 100 g of aluminum and chlorine gas?
2 Al(s)
100 g
+
3 Cl2(g)
1 mol Al 2 mol AlCl3
27 g Al 2 mol Al
xg
3.70 mol AlCl3
AlCl3
x g AlCl3 = 100 g Cl2
Cl2
2 AlCl3
100 g
x g AlCl3 = 100 g Al
Al

AlCl3
1 mol Cl2 2 mol AlCl3
.939 mol AlCl3
71 g Cl2 3 mol Cl2
1 mol Cl2 2 mol AlCl3 133.5 g AlCl3
x g AlCl3 = 100 g Cl2
= 125 g AlCl3
71 g Cl2 3 mol Cl2 1 mol AlCl3
Cl2 is limiting and only 125 g of AlCl3 is made
Limiting Reactants – Method 2

Begin by writing a correctly balanced chemical
equation
 Write down all quantitative values under
equation (include units)
 Convert ALL reactants to units of moles
 Divide by the coefficient in front of each
reactant
 The smallest value is the limiting reactant!

Limiting Reactants
How much product would be made if we begin
with 100 g of aluminum and chlorine gas?
2 Al(s)
100 g
Al
Cl2
AlCl3
AlCl3
+
3 Cl2(g)
100 g

2 AlCl3
xg
1 mol Al 2 mol AlCl3 133.5 g AlCl3
x g AlCl3 = 100 g Al
= 494 g AlCl3
1 mol AlCl3
27 g Al 2 mol Al
1 mol Cl2 2 mol AlCl3 133.5 g AlCl3
x g AlCl3 = 100 g Cl2
= 125 g AlCl3
71 g Cl2 3 mol Cl2 1 mol AlCl3
Cl2 is limiting and only 125 g of AlCl3 is made
C3H8
+
O2
------->
CO2
+
H2O
a) If you start with 14.8 g of C3H8 and 3.44 g of
O2, determine the limiting reagent
b) determine the number of moles of carbon
dioxide produced
c) determine the number of grams of H2O
produced
d) determine the number of grams of excess
reagent left
C 3H 8
+ 5 O2
-------> 3 CO2
a) O2
b) 0.065 mol CO2
c) 1.56 g H2O
d) 13.86 g C3H8
+
4 H 2O
Al2(SO3)3 + 6 NaOH ------> 3 Na2SO3
+ 2 Al(OH)3
a) If 10.0 g of Al2(SO3)3 is reacted with 10.0 g of NaOH,
determine the limiting reagent
b) Determine the number of moles of Al(OH)3
produced
c) Determine the number of grams of Na2SO3
produced
d) Determine the number of grams of excess reagent
left over in the reaction
a) Al2(SO3)3
b) 0.068 mol Al(OH)3
c) 12.85 g Na2SO3
d) 1.84 g NaOH