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Titrations. Viatmin C is both an acid and a reducing agent. One method of determining the amount of Vit C in a sample is to carry out the following reaction via a titration: 2H + (aq) + C 6 H 8 O 8 (aq) + 2Br 2 (aq) 4HBr(aq) + C 6 H 6 O 6 (aq) + 2H 2 O

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slide2

Viatmin C is both an acid and a reducing agent. One method of determining the amount of Vit C in a sample is to carry out the following reaction via a titration:

2H+(aq) + C6H8O8(aq) + 2Br2(aq)

4HBr(aq) + C6H6O6 (aq) + 2H2O

The vitamin C in a 1.00 g chewable tablet requires 28.28 mL of 0.102 M Br2 solution for titration to the equivalence point. How many grams of vitamin C (208 g/mol) are contained in the tablet?

mols C6H8O8

g C6H8O8

mols Br2

0.102 mols Br2  L

0.02828 L  1

1 mols C6H8O8 2 mol Br2

x

x

208g C6H8O8 mol C6H8O8

x

=

0.300 g C6H8O8

slide3

Viatmin C is both an acid and a reducing agent. One method of determining the amount of Vit C in a sample is to carry out the following reaction via a titration:

2H+(aq) + C6H8O8(aq) + 2Br2(aq)

4HBr(aq) + C6H6O6 (aq) + 2H2O

The vitamin C in a 1.00 g chewable tablet requires 28.28 mL of 0.102 M Br2 solution for titration to the equivalence point. How many grams of vitamin C (208 g/mol) are contained in the tablet?

mols C6H8O8

g C6H8O8

mols Br2

0.102 mols Br2  L

0.02828 L  1

1 mols C6H8O8 2 mol Br2

x

x

208g C6H8O8 mol C6H8O8

x

=

0.300 g C6H8O8

slide4

A 50.00 mL sample of solution containing Fe2+ is titrated with 0.0216 M KMnO4 solution. The solution required 20.65 mL of KMnO4 solution to oxidize all of the Fe2+ ions to Fe3+ by the reaction:

8H+(aq) + MnO41-(aq) + 5Fe2+(aq) 

Mn2+(aq) +5Fe3+(aq) + 4H2O

1

mols MnO41-

mols Fe2+

M Fe2+

mols KMnO4

2

0.0216 mols KMnO4 L

0.02065 L  1

1 mols MnO41- 1 mol KMnO4

x

x

5 mols Fe2+  1 mol MnO41-

1 0.0500 L

0.0446 M Fe2+

x

x

=

slide5

A 50.00 mL sample of solution containing Fe2+ is titrated with 0.0216 M KMnO4 solution. The solution required 20.65 mL of KMnO4 solution to oxidize all of the Fe2+ ions to Fe3+ by the reaction:

8H+(aq) + MnO41-(aq) + 5Fe2+(aq) 

Mn2+(aq) +5Fe3+(aq) + 4H2O

1

mols MnO41-

mols Fe2+

M Fe2+

mols KMnO4

2

0.0216 mols KMnO4 L

0.02065 L  1

1 mols MnO41- 1 mol KMnO4

x

x

5 mols Fe2+  1 mol MnO41-

1 0.0500 L

0.0446 M Fe2+

x

x

=

the limiting reagent
The Limiting Reagent
  • relative to the other reactant(s) of a chemical reaction, this reactant is present in less than the stoichiometrically equivalent amount

i.e. you have less than you need to fully react with other reactants.

  • determines,i.e. limits, the quantity of product(s) that will be obtained.
  • is totally consumed during the chemical reaction.

Reactants other than the limiting reagent are in excess (i.e. in excess of that amount required for stoi- chiometric equivalence with the limiting reagent). Some quantity of this (these) reactant(s) will remain after the reaction is complete.

slide7
Example
  • Desire
  • Provided with:

and

Which, or , is the

“limiting” supply?

slide8
CaC2(s) + 2H2O Ca(OH)2(aq) +C2H2(g)

How many grams of C2H2(g) will be formed from

the reaction of 24.0 g CaC2(s) and 18.0 g H2O ?

empirical formula
Empirical Formula

The simplest whole-numbered

(3)

ratio

(2)

of numbersofmolsofatoms

(1)

in one mol of a compound.

slide10

Mo(CO)x(s)

Mo(s)

+ xCO(g)

When a 2.200 g of Mo(CO)x is heated it decomposes producing 0.7809 g of Mo(s) and gaseous CO. The CO gas was found to occupy a volume of 1.2345 L at a temperature of 31.50 oC and a pressure of 751.1 mm Hg. What is the value of x?

R= 0.0821 L-atm/mol K

R = 62,400 mL-mm/mol K

slide11

A compound with the general formula CxHy was vaporized and, at 0.00 oC and 760 mm Hg, was found to have a density of 5.0996 g per L. In a separate determination the elemental composition of the compound was found to be 84.118 % C and 15.882 % H.

(1). Calculate the molar mass of the compound.

(2). Calculate the empirical formula of the compound

(3). What is the molecular formula of the compound?

R= 0.0821 L-atm/mol K

R = 62,400 mL-mm/mol K

slide12

How many grams of hydrogen gas are produced when 18 g C reacts with 27 g H2O?

C(s) + 2 H2O(l) CO2 + 2 H2(g)

slide13

C(s) + 2H2O  CO2(aq) +2H2(g)

From the balanced chemical equation:

1 mol C0.50 mols C

or

2 mols H2O mol H2O

Provided:

mol C

18.0g Cx  = 1.5 mol C

12.0 g C

mol H2O

27g H2O x  = 1.5 mol H2O

18.0 g H2O

Compare:

From

ProvidedReaction stoichiometry

1 mol C0.50 mols C

vs

1 mols H2O 1 mol H2O

Conclusion:

H2O is in the limiting reagent,C is in excess

slide14

mol H2O(l) 2 mol H2

27.0 g H2 O(l) x  x 

18.0 g H2O(l) 2 mols H2O(l)

2.02 g H2

x

mol H2(g)

3.03 g H2

=

slide15

How many grams of precipitate will be formed when 308.6 mL of 0.324 M Al2(SO4)3 is poured into 432. mL of 1.157 M NaOH?

Al2(SO4)3(aq)

+

6

NaOH

3

Na2SO4

(aq)

+

2

Al(OH)3

(s)

slide16
How many grams of N2O(g) will be produced when 14.0 g N2(g) reacts with 30.0 g H2O(g)?

N2(g) + H2O(g) N2O(g) + NH3(l)

How many grams H2(g) will be formed when

2.16 g Al react with 2.92 g HCl (in aqueous

solution)?

2 Al(s) + 6HCl(aq)  2 AlCl3(aq) + 3H2(g)

slide17
Al4C3(s) + H2O Al(OH)3(s) + CH4(g)

How many grams of CH4(g) will be formed from

the reaction of 14.4 g Al4C3(s) and 18.0 g H2O ?

slide18

Al4C3(s) + H2O  Al(OH)3(s) + CH4(g)

Which is the limiting reagent?

From the balanced chemical reaction:

1 mol Al4C3(s) 0.0833 mol Al4C3(s)

=

12 mols H2O 1 mols H2O

Provided:

mol Al4C3

14.4 g All4C3(s)x  = 0.100 mol Al4C3

144 g Al4C3

mol H2O

18g H2O x  = 1.0 mol H2O

18.0 g H2O

Compare:

From

ProvidedReaction stoichiometry

0.100 mol Al4C30.0833 mol Al4C3

vs

1.0 mol H2O 1.0 mol H2O

Al4C3(s) is in excess, , H2Ois the limiting reagent

slide19
Quantity of the product that will be obtained will be determined by the quantity of the limiting reagent provided.

1 mol H2O(l) 3 mol CH4 16.0 g CH4

18.0 gH2O(l) x  x  x 

18.0 g H2O (l) 12 mols H2O (l) mol CH4

= 4.0 g CH4

slide20
Which is the limiting reagent?

From the balanced chemical reaction:

1 mol Al4C3(s)

12 mols H2O

Now, how many molsAl4C3(s) are needed to react

with the number of mols of H2O provided?

1 mol H2O(s) 1 molsAl4C3(s)

18.0 gH2O(s)x x =

18 g H2O(s) 12 mols H2O (l)

= 0.083 molsAl4C3(s) (needed)

18.0gH2Orequires 0.083 mols Al4C3(s) for complete reaction.

14. 4 g Al4C3(s) is provided:

1 mol Al4C3(s)O

14.4 g Al4C3(s) x = 0.10 mols Al4C3(s)

144 g Al4C3(s)

0.10 mols Al4C3(s) (provided) > 0.083 mols Al4C3(s) H2O (needed )

Conclusion:

Al4C3(s) is in excess, H2O is the limiting reagent

slide21
Quantity of the product that will be obtained will be determined by the quantity of the limiting reagent provided.

1 mol H2O(l) 3 mol CH4 16.0 g CH4

18.0 gH2O(l) x —————— x ————— x —————

18.0 g H2O (l) 12 mols H2O (l) mol CH4

= 4.0 g CH4

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