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) + 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 Br2 mols C6H8O8

g C6H8O8

0.102 mols Br2

L

x0.02828 L

1x

1 mols C6H8O8

2 mol Br2

x 208 g C6H8O8

mol C6H8O8

= 0.300 g C6H8O8

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 Br2 mols C6H8O8

g C6H8O8

0.102 mols Br2

L

x0.02828 L

1x

1 mols C6H8O8

2 mol Br2

x 208 g C6H8O8

mol C6H8O8

= 0.300 g C6H8O8

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 KMnO4

mols MnO4

1-

mols Fe2+

M Fe2+

0.0216 mols KMnO4

L

x0.02065 L

1x

1 mols MnO41-

1 mol KMnO4

x 5 mols Fe2+

1 mol MnO4

1-

1 0.0500 L

0.0446 M Fe2+ x =

2

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 KMnO4

mols MnO4

1-

mols Fe2+

M Fe2+

0.0216 mols KMnO4

L

x0.02065 L

1x

1 mols MnO41-

1 mol KMnO4

x 5 mols Fe2+

1 mol MnO4

1-

1 0.0500 L

0.0446 M Fe2+ x =

2

The Limiting Reagent

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.

is totally consumed during the chemical reaction.

determines,i.e. limits, the quantity of product(s) that will be obtained.

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.

Example

• Desire

• Provided with:

and

Which, or , is the

“limiting” supply?

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

The simplest whole-numbered

(3)ratio

(2) of numbers of mols of

atoms (1)

in one mol of a compound.

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

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

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

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

From the balanced chemical equation: 1 mol C 0.50 mols C

or

2 mols H2O mol H2O

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

Conclusion: H2O is in the limiting reagent, C is in excess

From Provided Reaction stoichiometry 1 mol C 0.50 mols C vs 1 mols H2O 1 mol H2O

Compare:

mol H2O27g H2O x = 1.5 mol H2O 18.0 g H2O

mol C18.0g C x = 1.5 mol C 12.0 g C

Provided:

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

Al2(SO4)3(aq)

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?

+ NaOH

Al(OH)3Na2SO4 (aq) + (s)23

6

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)

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 ?

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 Provided Reaction stoichiometry

0.100 mol Al4C3 0.0833 mol Al4C3

vs

1.0 mol H2O 1.0 mol H2O

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

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

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 g H2O (l) x x x

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

= 4.0 g CH4

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 g H2O(s) x x =

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

= 0.083 molsAl4C3(s) (needed)

18.0g H2O requires 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

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 g H2O (l) x —————— x ————— x —————

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

= 4.0 g CH4