Transcript Document

LIMITING REACTANT
The reactant that gives the least number of
product moles “limits” the reaction.
To understand this concept, let’s suppose you were an elf working for
Santa Claus and your job was to make candy canes. You take one red
stick and one white stick then twist them around to make one candy
cane. The ratio of red to white is 1:1. Time is ticking and you find that
you have 24 red stick and 17 white sticks left. What is the maximum
number of candy canes you can make?
The answer is 17 candy canes! The white sticks “limit” the
amount of product you could make. In chemistry we do not
use sticks but *MOLES* to determine which starting
material will limit the maximum amount of product that can
be produced in a chemical reaction.
LIMITING REACTANT
a) Sodium metal reacts with oxygen to produce sodium
oxide. If 5.00 g of sodium reacted with 5.00 grams of
oxygen, how many grams of product is formed?
4 Na (s) + O2(g)  2 Na2O (s)
Start with what is given, calculate the amount of product that can be
theoretically made but do it twice (once for each reactant):
5.00g Na (1 mole Na) ( 2 mole Na2O)( 62 g Na2O) = 6.74 g of Na2O
23 g Na
4 mole Na
1 mol Na O
2
5.00g O2 (1 mole O2) ( 2 mole Na2O)( 62 g Na2O) = 19.38
g of Na
Wrong
answer
2O
32 g O
2
1 mole O
2
1 mol Na O
2
Notice you can not have two different masses produced for the same
product in one reaction vessel! So in this case, Na (sodium) “limits”
how much sodium oxide is produced. The correct answer is 6.74 g of
sodium oxide.
LIMITING REACTANT
b) How much oxygen was used in this reaction and how
much of each reactant was leftover (in excess)?
4 Na (s) + O2(g)  2 Na2O (s)
There are two methods used to answer this question. The Law of
Conservation of mass and Stoichiometry.
The amount of O2 used to make 6.74 g of Na2O is calculated by:
5.00g Na (1 mole Na) ( 1 mole O2)( 32 g O2) = 1.74 g of O2 was used
23 g Na
4 mole Na
1 mol O
2
Or use the Law of Conservation of mass:
Mass of product (6.74 g) – mass of limiting reactant (5.00 g) = mass of other
reactant, in this case oxygen (1.74 g).
The amount of oxygen (O2) leftover can be calculated by subtracting the starting
mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g
The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero),
since it was the limiting reactant and was completely consumed in the reaction.
LIMITING REACTANT
How many grams of solid are formed when 10.0 g
of lead reacts with 10.0 g of phosphoric acid?
3 Pb+2 H3PO4  Pb3(PO4)2 (s) + 3 H2 (g)
Start with what is given but do it twice:
10.0g Pb (1 mole Pb) ( 1 mole Pb3(PO4)2)(811 g Pb3(PO4)2) = 13.1 g Pb3(PO4)2
207 g
3 mole Pb
1 mole Pb (PO )
3
42
Wrong
answer
10.0g H3PO4 (1 mole H3PO4)( 1 mol Pb3(PO4)2)(811 g Pb3(PO4)2)=41.4 g Pb3(PO4)2
98 g
2 mole H PO 1 mole Pb (PO )
3 4
3
42
You can not have two different answers for one question so in this
case lead “limits” how much lead(II) phosphate that can be
produced. The correct answer is 13.1 g.
PRACTICE PROBLEM #19
1. How much AgCl product will be produced if 100.00 g of
BaCl2 reacted with excess AgNO3?
137.57 g
2. How many moles of carbon dioxide could be produced
from 220.0 g of C2H2 and 545.0 g of O2?
13.63 mol
3. How many grams of CO2 can be produced by the reaction
of 35.5 grams of C2H2 and 45.9 grams of O2?
50.5 g
4. In the reaction between CH4 and O2, if 25.0 g of CO2 are
produced, what is the minimum amount of each reactant
needed?
9.09 g of CH & 36.4 g of O
4
2
5. Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag.
When 10.0 g of copper was reacted with 60.0 g of silver
nitrate solution, 30.0 g of silver was obtained. What is the
88.3%
percent yield of silver obtained?
GROUP STUDY PROBLEM #19
______1. Which reactant will produce the least amount of AgCl
product if reacted with AgNO3?
a)100.00 g BaCl2
b) 400.0 g NaCl
c) 200.0 g CsCl
______2. How many moles of CO2 can be produced by the reaction of
5.0 grams of C2H4 and 12.0 grams of O2?
______3. How many grams of carbon dioxide could be produced from
2.0 g of C2H4 and 5.0 g of O2?
______4. In the reaction between CH4 and O2, if 18.0 g of CO2 are
produced, how many grams of water are produced?
______5. Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag.
When 50.0 g of copper was reacted with 300.0 g of silver nitrate
solution, 149 g of silver was obtained. What is the percent yield of
silver obtained?