Sample Q 4. 2007 Examination

Sample Q 4. 2007 Examination• DIRECTIONS: For each of the following three reactions, in part

(i) write a BALANCED equation and in part (ii) answer the question about the reaction. Coefficients should be in terms of lowest whole numbers. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solutions as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction.

Q 4. Example

• Example: A strip of magnesium is added to a solution of silver nitrate

• Mg + 2 Ag + Mg 2+ + 2 Ag

• Which substance is oxidized in the reaction?

Magnesium metal (Mg)

Sample 2006 recast for 2007:• 4a. Solid potassium chlorate is strongly heated resulting in a

change in the oxidation numbers of both chlorine and oxygen.

(i) 2 KClO3 2 KCl + 3 O2

– (ii) What is the oxidation number of chlorine before and after the reaction?

Chlorine has an oxidation number of +5 in KClO3 and -1 in KCl

Sample 2006 recast for 2007:

• 4b. Solid silver chloride is added to a solution of concentrated hydrochloric acid forming a complex ion.

(i) AgCl + Cl- [AgCl2 ] -

• (ii) Which species acts as a Lewis base in the reaction? Explain.

The chloride ion acts as a Lewis base in the reaction because it donates an electron pair in the reaction.

Sample 2006 recast for 2007:

• 4c. A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide.

(i) HC2H3O2 + OH- H2O + C2H3O2-

• (ii) Explain why a mixture of equal volumes of equimolar solutions of ethanoic acid and barium hydroxide is basic.

In the mixture there are initially twice as many moles of hydroxide ions as moles of acid; since they react in a 1:1 ratio, there is an excess of hydroxide ions, leading to the basic solution.

Sample 2006 recast for 2007:

• 4d. Ammonia gas is bubbled into a solution of hydrofluoric acid.

(i) NH3 + HF NH4+ + F-

• • (ii) Identify a conjugate acid-base pair in the reaction.

NH3 (base) and NH4+ (acid) OR HF (acid) and F-(base).

Sample 2006 recast for 2007:

• 4e. Zinc metal is placed in a solution of copper(II) sulfate.

• (i) Zn + Cu2+ Zn2+ + Cu •

• (ii) Describe the change in color of the solution that occurs as the reaction proceeds.

The blue color of the solution due to the presence of the hydrated copper(II) ion fades as the copper(II) ion reacts and the colorless hydrated zinc(II) ion forms.

Sample 2006 recast for 2007:

• 4f. Hydrogen phosphide (phosphine) gas is added to boron trichloride gas.

(i) PH3 + BCl3 H3PBCl3

• (ii) Which species acts as a Lewis acid in the reaction? Explain.

BCl3 , because it accepts the non-bonded pair of electrons of the phosphorous atom in PH3

Sample 2006 recast for 2007:

• 4g. A solution of nickel(II) bromide is added to a solution of potassium hydroxide.

(i) Ni2+ + 2 OH- Ni(OH)2

• (ii) Identify the spectator ions in the reaction mixture.

Spectator ions are the bromide ion (Br-) and the potassium ion (K+)

Sample 2006 recast for 2007:

• 4h. Hexane is combusted in air.

(i) 2 C6H14 + 19 O2 12 CO2 + 14 H2O

• (ii) When one molecule of hexane is completely combusted, how many molecules of products are formed?

1 molecule of hexane produces 13 molecules of products

2 H2 + O2 -----> 2 H2O

You need a balancedequation and you WILL

work with moles.

Mass relationships in chemical reactions

• Stoichiometry - The calculation of quantities of reactants and products in a chemical reaction.

Stoichiometry, General steps.1 Write a balanced chemical equation.

3 Use the coefficients from the chemical equation to convert moles of the given substance to moles of the unknown substance.

2 Calculate the moles of the given quantity,using molar mass, molarity or molarvolume of a gas at STP as conversionfactors.

4 Convert back to mass, or any other unit of measurement, if needed.

Stoichiometry calculations•Red phosphorus, P4, reacts explosively with KClO3 when struck with a hammer. How many grams of KClO3(s) would be needed to react completely with 0.30 g of P4(s)?

1 Write a balanced chemical equation.P4(s) + 2KClO3(s) 2P2O3(g) + 2KCl(s)

By predicting products, metal chlorates decompose into metal chlorides and oxygen gas. So….. KClO3(s) forms KCl and O2

P4 and O2 can combine to form PxOy. The most stable oxidation state of P is 3+, so P2O3 is the most likely product.

Stoichiometry calculations•Red phosphorus, P4, reacts explosively with KClO3 when struck with a hammer. How many grams of KClO3(s) would be needed to react completely with 0.30 g of P4(s)?2 Calculate the moles of the given quantity,

using molar mass, molarity or molarvolume of a gas at STP as conversionfactors.

X moles P4 = 0.30 g P4 ( 1mole P4 )= 0.00242 mole 124 g P4

Stoichiometry calculations

•X moles KClO3 = 0.00242 mol P4 ( 2 mole KClO3 ) =

• 1 mole P4

• 0.00484 mol KClO3

3 Use the coefficients from the chemical equation to convert moles of the given substance to moles of the unknown substance.

Stoichiometry calculations

•x grams KClO3 = 0.00484 mol KClO3( 122.5 g KClO3 )

• 1 mole KClO3

• = 0.593 grams KClO3

4 Convert back to mass, or any other unit of measurement, if needed.

Volume calculations

• What volume of P2O3 gas will be formed from the complete reaction of 0.30 g P4(s) with the KClO3(s) at STP?

– The same balanced equation applies:

• P4(s) + 2KClO3(s) 2P2O3(g) + 2KCl(s)

X L P2O3 = 0.30g P4 (1mol P4) 124 g P4

(2 mol P2O3) 1 mol P4

(22.4 L P2O3) 1 mol P2O3

= 0.108 L P2O3

Stoichiometry calculations

• How many grams of hydrogen will be produced if 1.00 grams of calcium is added to an excess of hydrochloric acid?

Note:•We produce one H2 for each calcium.• There is an excess of HCl so we have all we need.

Ca(s) + + CaCl2(aq)HCl(aq) H2(g)2

Stoichiometry calculations

• Moles Ca = grams Ca / FM Ca

» = 1.00 g

» = 0.0250 mol Ca

1 mol40.08 g

2HCl + Ca ____> CaCl2 + H2

First - Determine the number of moles of calcium available for the reaction.

Stoichiometry calculations

» 2HCl + Ca _____> CaCl2 + H2

» 1.00 g Ca = 0.0250 mol Ca

• According to the chemical equation, we get one mole of H2 for each mole of Ca.

• So we will make 0.0250 moles of H2.grams H2 produced = moles x FW H2

= 0.0250 mol x 2.016 g/mol

= 0.0504 grams H2

Stoichiometry calculations

•How many mL of hydrogen gas would be produced at STP?

• (Remember we will make 0.0250 moles of H2.)mL H2 produced = moles x molar volume at STP

= 0.0250 mol x 22,400 mL/mol

= 560. mL H2

Stoichiometry calculations

• OK, so how many grams of CaCl2 were made?

» 2HCl + Ca _____

> CaCl2 + H2

» 1.00 g Ca = 0.0250 mol Ca

– We would also make 0.0250 moles of CaCl2.g CaCl2 = 0.0250 mol x FM CaCl2

= 0.0250 mol x 110.98 g / mol CaCl2

= 2.77 g CaCl2

Stoichiometry calculations

• What is the minimum volume of 0.50 M HCl(aq) that is needed to completely react with the calcium?

» 2HCl + Ca _____

> CaCl2 + H2

» 1.00 g Ca = 0.0250 mol CamL HCl = 0.0250 mol Ca x mole ratio molarity

= 0.0250 mol Ca x x2 mol HCl1 mol Ca

1000 mL HCl0.50 mol HCl

= 100. mL HCl

Limiting reactant

• In the last example, we had HCl in excess.

• The reaction stopped when we ran out of Ca.

• Ca is considered the limiting reactant.

• Limiting reactant - the material that is in the shortest supply based on a balanced chemical equation.

Excess Problem Example• Which is the limiting factor if you have a

reaction between 5.0 g of hydrogen and 10 g oxygen?

1 mol32.0 g

1 mol2.0 g

Balanced Chemical Reaction

2H2 + O2 ________> 2H2O

You need 2 moles of H2 for each mole of O2.

Moles of H2 = 5 g x = 2.5 mol

Moles of O2 = 10 g x = 0.31 mol

Excess Problem Example• Balanced Chemical Reaction

• 2H2 + O2 2H2O

• You need 2 moles of H2 for each mol of O2

• You have 2.5 moles of H2 and 0.31 mol of O2

• Need a ratio of 2:1 but we have a ratio of 2.5 : 0.31 or 8.3 : 1.

– Hydrogen is in excess and oxygen is the– limiting reactant.

The mole-gas volume relationship: Avogadro’s LawAt a given temperature and pressure, the volumes of gases

which react are ratios of small whole numbers. Avogadro’s Law states that equal volumes of these gases will contain the same number of moles of each gas.

Stoichiometry Problem Solving• First step is to determine the given quantities

and what unknown information you are required to find.

• Next, determine the number of moles of the given quantities present.

• Write a balanced chemical equation for the reaction.

Stoichiometry Problem Solving

• Use the coefficient ratio from the equation to determine the moles of the unknown substance. (If quantities of both reactants are given, you must check to see if an excess occurs.)

• Convert the moles of the unknown substance into the desired units.

Stoichiometry Example Problem

• Given quantities• 2.50 grams of Ca(OH) 2

• Unknown quantities• volume of 0.100 M HCl needed

Determine the volume of 0.100 M HCl that must be added to completely react with 2.50 grams of Ca(OH)2

Stoichiometry Example Problem

• X moles Ca(OH)2 =

• 2.50g Ca(OH)2 x 1 mole Ca(OH)2

• 74.1 g Ca(OH)2

• = 0.0337 mole Ca(OH)2

Determine the volume of 0.100 M HCl that must be added to completely react with 2.50 grams of Ca(OH)2

Mole calculations• The balanced equation shows the reacting ratio

between reactants and products.

• Ca(OH)2 + 2HCl CaCl2 + 2H2O

– You need twice as many moles of HCl as the 0.0337 mole Ca(OH)2 present, which equals 0.0674 mole HCl.

– The number of moles can be converted into mass quantities, volumes of solutions or gases, or number of particles (molecules or ions).

Mass calculations

• X mL HCl =

• 0.0674 mole HCl x 1000 mL HCl• 0.100 mole

• = 674 mL HCl needed

Limiting reactant• In the last example, we needed 674 mL HCl to

react completely. Any amount of HCl less than this results in the reaction stopping before all of the Ca(OH)2 is consumed.

• Reaction stops when the HCl runs out.

• HCl is considered the limiting reactant.

• Limiting reactant - the material that is in the shortest supply based on a balanced chemical equation.

Example• Determine the mass of CaCl2 that can be

produced if only 500. mL of 0.100 M HCl was added to 2.50 g of Ca(OH)2. How many grams of excess Ca(OH)2 remain unchanged?

• Balanced Chemical Reaction• Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O0.100 mol

1000 mL

1mol CaCl22 mol HCl

111g CaCl2

1 mol CaCl2

74.1g Ca(OH)2

1 mol Ca(OH)2

moles of HCl = 500. mL x = 0.0500 mol HCl

#g of CaCl2=0.0500 mol HCl x x

#g excess Ca(OH)2=[0.0337mol added - 0.0250mol reacts]

x

= 2.78 g CaCl2

= 0.645 g Ca(OH)2 remains

Theoretical, actual and percent yields

• Theoretical yield - The amount of product that should be formed according to the chemical reaction.

• Actual yield - The amount of product actually formed.

• Percent yield - Ratio of actual to theoretical yield, as a %.

• Quantitative reaction - When the percent yield equals 100%.

Yield• Less product is often produced than expected.

• Possible reasons• A reactant may be impure.• Some product is lost mechanically since the

product must be handled to be measured.• The reactants may undergo unexpected reactions

- side reactions.• No reaction truly has a 100% yield due to the

limitations of equilibrium.

Percent yield• The amount of product actually formed

divided by the amount of product calculated to be formed, times 100.

• % yield = x 100

Actual yieldTheoretical yield

In order to determine % yield, you must be able to recover and measure all of the product in a pure form.

% Yield example• Example. The final step in the production of

aspirin is the reaction of salicylic acid with acetic anhydride.

• 48.6 g of aspirin is produced when 50.0 g of salicylic acid and an excess of acetic anhydride are reacted. What is the % yield?

HOC6H4COOH(s) + (CH3CO)2O(l)

salicylic acid acetic anhydride

CH3OC6H4COOH(s) + CH3COOH(l)

aspirin acetic acid

% Yield example

• Number of moles of salicylic acid used:

One mole of aspirin should be produced for each mole of salicylic acid consumed. Number of grams of aspirin that should have been produced -- theoretical yield:

(0.362 mol aspirin)( 180 g/mol) = 65.2 g aspirin

= 0.362 mole of salicylic acid 1 mol138 g50.0 g

% Yield example• % Yield for this reaction

• Theoretical yield = 65.2 g• Actual yield = 48.6 g

• % Yield = x 100

• = 74.5%

• Yields less than 100% are very common in industrial processes.

48.6 g65.2 g

Solution stoichiometry• Extension of earlier stoichiometry problems.

• First step is to determine the number of moles based on solution concentration and volume.

• Final step is to convert back to volume or concentration as required by the problem.

• You still need a balanced equation and must use the coefficients for working the problem.

Solution stoichiometry example

• Determine the volume of 0.100 M H2SO4 that must be added to completely react with 250 ml of 2.50 M NaOH

• Balanced chemical equation

• H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + H2O (l)

• The first step is to determine how many moles of NaOH we have.

Solution stoichiometry example• We have 250 ml of a 2.50 M solution.

• molNaOH = 0.250 L x 2.50 mol/L

• = 0.625 molNaOH

• From the balanced chemical equation, we know that we need two moles of NaOH for each mole of H2SO4.

• That means we need 0.312 mol H2SO4.

Solution stoichiometry example• Now we can determine what volume of our

0.100 M H2SO4 solution is required.

• L = mol H2SO4 / M H2SO4

• = 0.312 mol x

• = 3.12 L

1 L0.100 mol( )