# Stoichiometry.

## Presentation on theme: "Stoichiometry."— Presentation transcript:

Stoichiometry

S t o i c h i o m e t r y A q u a n t i t a t i v e s t u d y o f c h e m i c a l c h a n g e s

M a s s - M a s s P r o b l e m s G i v e n a c e r t a i n a m o u n t o f r e a c t a n t , h o w m u c h p r o d u c t c a n b e f o r m e d ?

M a s s - M a s s P r o b l e m s T h e m o l e r a t i o f r o m
t h e c o e f f i c i e n t s i n a b a l a n c e d e q u a t i o n i s u s e d t o c h a n g e f r o m o n e s u b s t a n c e t o a n o t h e r .

M o l e F r a c t i o n FeS + 2HCl H2S + FeCl2 1 mole FeS = 1 mole H2S

M o l e F r a c t i o n FeS + 2HCl H2S + FeCl2 2 mole HCl = 1 mole H2S

Mass-Mass Steps: 1. Write the balanced equation.

Mass-Mass Steps: 2. Put given mass on a factor-label form.

3. Convert mass of reactant to moles of reactant.
Mass-Mass Steps: 3. Convert mass of reactant to moles of reactant.

Mass-Mass Steps: 4. Convert moles of reactant to moles of product.

Mass-Mass Steps: 5. Convert moles of product to grams of product.

Mass-Mass Steps: 6. Pick up the calculator and do the math.

M a s s - M a s s p r o b l e m s u s u a l l y l o o k l i k e t h i s :

t h e a i r w i l l c o m b i n e w i t h b o t h t h e i r o n
I f i r o n p y r i t e , F e S 2 , i s n o t r e m o v e d f r o m c o a l , o x y g e n f r o m t h e a i r w i l l c o m b i n e w i t h b o t h t h e i r o n a n d t h e s u l f u r a s c o a l b u r n s.

W r i t e a b a l a n c e d e q u a t i o n s h o w i n g t h e f o r m a t i o n o f i r o n ( I I I ) o x i d e a n d s u l f u r d i o x i d e.

N o w t h e m a s s - m a s s p a r t :
I f a f u r n a c e b u r n s a n a m o u n t o f c o a l c o n t a i n i n g g o f F e S 2 , h o w m u c h s u l f u r d i o x i d e i s p r o d u c e d ?

M a s s - M a s s P r o b l e m s S t e p 1:
W r i t e t h e b a l a n c e d e q u a t i o n f o r t h e r e a c t i o n .

F e S O F e 2 O S O 2

B a l a n c e t h e e q u a t i o n : D i a t o m i c m o l e c u l e s M e t a l s N o n m e t a l s O x y g e n H y d r o g e n R e c o u n t a l l a t o m s R e d u c e , i f n e e d e d

4 F e S O F e 2 O S O 2

M a s s - M a s s P r o b l e m s S t e p 2:
W r i t e w h a t i s g i v e n .

g F e S 2 W r i t e w h a t i s g i v e n

M a s s - M a s s P r o b l e m s S t e p 3:
C o n v e r t m a s s o f r e a c t a n t t o m o l e s o f r e a c t a n t .

1 0 0 g F e S 2 1 m o l e F e S 2 1 2 0 g F e S 2 formula mass: Fe = 1 X 56 = 56 S = 2 X 32 = 64 120 C o n v e r t g r a m s t o m o l e s

M a s s - M a s s P r o b l e m s S t e p 4:
U s e c o e f f i c i e n t s o f r e a c t a n t a n d p r o d u c t i n t h e b a l a n c e d e q u a t i o n

R e a c t a n t 4 F e S O F e 2 O S O 2 P r o d u c t

M a s s - M a s s P r o b l e m s S t e p 4:
C o n v e r t m o l e s o f r e a c t a n t t o m o l e s o f p r o d u c t .

1 m o l e F e S 2 1 0 0 g F e S 2 2 m o l e S O 2 1 m o l e F e S 2
C h a n g e s u b s t a n c e s

M a s s - M a s s P r o b l e m s S t e p 5:
C o n v e r t m o l e s o f p r o d u c t t o g r a m s o f p r o d u c t .

1 m o l e F e S 2 1 0 0 g F e S 2 2 m o l e S O 2 6 4 g S O 2
1 m o l e F e S m o l e S O 2 C o n v e r t m o l e s t o g r a m s

M a s s - M a s s P r o b l e m s S t e p 6:
U s e c a l c u l a t o r t o d o t h e m a t h .

= 1 0 7 g S O 2 100 g FeS2 1 mole FeS2 2 mole SO2 64 g SO2
D o t h e m a t h

R e m e m b e r : F a c t o r - l a b e l c a n c e l s u n i t s
100 g FeS2 1 mole FeS2 2 mole SO g SO2 120 g FeS mole FeS2 1 mole SO2 R e m e m b e r : F a c t o r - l a b e l c a n c e l s u n i t s

P r a c t i c e

Problem #1

Problem #2

Problem #3

More P r a c t i c e

How much silver is obtained from the decomposition of 100 grams of silver oxide?

2Ag2O --> 4Ag + O2

100g Ag2O 1 mole Ag2O 4 mole Ag 108g Ag 232g 2 mole 1 mole Ag Ag2O Ag2O
Ag - 2 X 108 = 216 O X 16 = 16 232 93g Ag

50g of iron (II) sulfide react with an excess of HCl in a double displacement reaction. How many grams of hydrogen sulfide are produced?

FeS + 2HCl --> H2S + FeCl2 50g FeS 1 mole FeS 1 mole H2S 34g H2S 88g FeS 1 mole FeS 1 mole FeS

How much ammonium hydroxide is produced when 25g of ammonium phosphate react with an excess of barium hydroxide in a double displacement reaction?

2(NH4)3PO4 + 3Ba(OH)2 --> 6NH4OH + Ba3(PO4)2 25g mole mole g (NH4)3PO4 (NH4)3PO4 NH4OH NH4OH g mole mole (NH4)3PO4 (NH4)3PO4 NH4OH 18g NH4OH

Calcium and aluminum chloride combine in a single displacement reaction. Assuming a sufficient amount of calcium is used, how much aluminum is produced from 10g of aluminum chloride?

3Ca + 2AlCl3 --> 2Al + 3CaCl2 10g AlCl3 1 mole AlCl3 2 mole Al g Al g AlCl mole AlCl3 1 mole Al 2g Al

Limiting Reactant

The reactant that is completely consumed in a reaction

the reactant that remains after the reaction stops
Excess reactant: the reactant that remains after the reaction stops

The amount of product formed depends on the limiting reactant

Steps used in solving a limiting reactant problem

1. Write a balanced equation

2. Convert BOTH reactant quantities to moles

3. Determine moles of product formed by each reactant

4. The one producing the LEAST amount is the limiting reactant

Finish the calculation using the limiting reactant
5. Finish the calculation using the limiting reactant

Sample limiting reactant problem
What mass of water can be produced by 4g of H2 reacting with 16g of O2?

Write a balanced equation
Sample limiting reactant problem 1. Write a balanced equation for the reaction.

2 H2 + O H2O 2. Convert both reactant quantities to moles.

Using the mole ratio from the equation, determine the
3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant.

4. Oxygen produces the least amount of water.
16 grams of O2 cannot produce as much water as 4 grams of H2. In other words, 16 grams of O2 will be used up in the reaction before 4 grams of H2.

Oxygen is the "limiting" reactant. Use oxygen to finish the calculation of product amount.

Complete the problem by
5. Complete the problem by converting moles of H2O to mass of H2O.

Limiting reactant practice problem:
75 grams of calcium oxide react with 130 grams of hydrochloric acid to produce a salt and water. What is the limiting reactant?

Limiting reactant practice problem:
CaO + HCl Salt + H2O

Limiting reactant practice problem:
CaO + HCl CaCl2 + H2O Balance the equation

Limiting reactant practice problem:
CaO + 2HCl CaCl2 + H2O Convert both reactants to moles

Convert both reactants to moles of product
75 g CaO 1 mole CaO 56 g CaO Ca - 1 X 40 = 40 O - 1 X 16 = 16 56 130 g HCl 1 mole HCl H - 1 X 1 = 1 Cl - 1 X 35 = 35 36 g HCl 36 Convert both reactants to moles of product

Whichever produces the LEAST amount of product is the
75 g CaO 1 mole CaO 1 mole CaCl2 56 g CaO mole CaO 130 g HCl 1 mole HCl 1 mole CaCl2 36 g HCl mole HCl Whichever produces the LEAST amount of product is the limiting reactant.

CaO produces 1.3 mole CaCl2 HCl produces 1.8 mole CaCl2
75 g CaO 1 mole CaO 1 mole CaCl2 56 g CaO mole CaO 130 g HCl 1 mole HCl 1 mole CaCl2 36 g HCl mole HCl CaO produces 1.3 mole CaCl2 HCl produces 1.8 mole CaCl2

CaO is the limiting reactant.
75 g CaO 1 mole CaO 1 mole CaCl2 56 g CaO mole CaO 130 g HCl 1 mole HCl 1 mole CaCl2 36 g HCl mole HCl CaO produces 1.3 mole CaCl2 CaO is the limiting reactant.

Homework Problem: How much aluminum oxide is produced when
46.5 g of Al react with 165.37g of MnO?

Homework Problem 2 Al + 3 MnO Mn + Al2O3

Homework Problem 46.5g Al 1 mole Al 1 mole Al2O3 27g Al 2 moles Al
0.86 46.5g Al 1 mole Al 1 mole Al2O3 27g Al moles Al Al - 1 X 27 = 27 0.78 165.37g MnO 1 mole MnO 1 mole Al2O3 Mn - 1 X 55 = 55 O - 1 X 16 = 16 71g MnO mole MnO 71

Homework Problem 165.37g MnO 1 mole MnO 1 mole Al2O3 102g Al2O3

5 grams of copper metal react with a solution containing
#1 5 grams of copper metal react with a solution containing 20 grams of silver nitrate to produce copper (II) nitrate and silver. A. What is the limiting reactant? B. How many grams of Ag are produced?

1A. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag 5g Cu 1 mole Cu 2 mole Ag
0.15 64g Cu mole Cu 0.12 20g AgNO3 1 mole AgNO3 2 mole Ag 170 g AgNO mole AgNO3 silver nitrate is the limiting reactant

1B. Cu AgNO Cu(NO3) Ag 20g AgNO3 1 mole AgNO3 2 mole Ag g Ag 170 g AgNO mole AgNO3 1 mole Ag 12.7g Ag

#2 A solution containing 20 g of sodium
sulfite reacts with 7 ml of phosphoric acid. The concentration of the acid solution is such that there are 1.83 g of H3PO4 per ml of solution. A. How many g of water are produced? B. How many moles of Na3PO4 are produced? C. How many g of SO2 are produced?

3Na2SO3 + 2H3PO Na3PO4 + 3SO2 + 3H2O 20g Na2SO3 1 mole Na2O3 3 mole H2O 0.159 126g Na2SO3 3 mole Na2SO3 12.8g H3PO4 1 mole H3PO4 3 mole H2O 0.196 98g H3PO mole H3PO4 1.83 g ml = 12.8 g 1 ml

2A. 3Na2SO3 + 2H3PO Na3PO4 + 3SO2 + 3H2O 20g Na2SO3 1 mole Na2O3 3 mole H2O g H2O 126g Na2SO3 3 mole Na2SO3 1 mole H2O 2.9 g H2O Produced

2B. 3Na2SO3 + 2H3PO4 2Na3PO4 + 3SO2 + 3H2O 17.4 g Na3PO4 Produced
20g Na2SO3 1 mole Na2O3 2 mole Na3PO g Na3PO4 126g Na2SO3 3 mole Na2SO mole Na3PO4 17.4 g Na3PO4 Produced

2C. 3Na2SO3 + 2H3PO4 2Na3PO4 + 3SO2 + 3H2O 10.2 g SO2 Produced
20g Na2SO3 1 mole Na2O3 3 mole SO g SO2 126g Na2SO3 3 mole Na2SO mole SO2 10.2 g SO2 Produced

Thirteenth Lab Copper and Silver Metals

END Stoichiometry