Balanced Chemical Equations

C + O₂                        CO₂ One mole of carbon is reacting with one mole of oxygen to form one mole of carbon dioxide. 2Al + 3O₂                       2Al₂O₃ 2 moles aluminium reacts with 3 moles of oxygen to form 2 moles of aluminium oxide. Mg + 2HCl                       MgCl₂ + H₂ One mole of magnesium is reacting with 2 moles of hydrochloric acid to form one moles of magnesium chloride and one mole of hydrogen gas.   Banner 4

CALCULATIONS FROM EQUATIONS

Calculate the mass of Magnesium oxide produced from 6 g of Magnesium when burned complete y in air ? STEP 1: Write the balanced chemical equation. 2Mg + O₂                        2MgO STEP2: Write known to the left and unknown to the right Mg = MgO STEP3: Write the moles relationship from the balanced chemical equation 2moles = 2 moles STEP4: Convert moles into mass 48g = 80g STEP5: Do the Maths. 48/48 = 1g                                                 80/48 = 1.66g 1g× 6 = 6g                                                 1.66g × 6 = 10g Calculate the mass of alumium needs to produce 306 g of Aluminium Oxide ? STEP 1: Write the balanced chemical equation. 4Al + 3O₂                        2Al₂O₃ STEP2: Write known to the left and unknown to the right 2Al₂O₃                         Al STEP3: Write the moles relationship from the balanced chemical equation 2moles = 4 moles STEP4: Convert moles into mass 204g = 108 STEP5: Do the Maths. 306g                       108/204 x 306 = 162g Banner 5

LIMITING REAGENTS

To work out the limiting reagent work out moles of all the reagents and then work from the balanced chemical equation which reagent is in excess and which is in limiting The reactant that get all used up completely is the limiting reagent. The other reagent which is present in greater quantity than required is in excess. 13.5 gm Aluminium reacts which 32 g of Oxygen

1. a) Work out the moles of Aluminium and Oxygen
2. b) Which reagent is limiting and which is in excess
3. c) Calculation the mass of aluminium oxide produced ?

4Al + 3O₂                        2Al₂O₃

13. Moles of Al = 13.5/27 = 0.5 moles

Moles of O₂ = 32/32 = 1 moles

1. 4 moles of Alumunium = 3 moles of oxygen

1 moles of alumnum = ¾ O₂ 0.5 mol of aluminium = ¾ x 0.5 = 0.375 So 0.375 moles oxygen is required hence oxygen is in excess and alumunium is limiting

1. Al = A₂O₃

4 moles = 2 moles

• 2/4

0.5 – 2/4 x 0.5 =0.25 x 102 =25.5g Baneer 6

PERCENTAGE YIELD

% yield = observed mass/expected mass x 100 = Actual yield/Theoretical yield x 100 When 28 g of nitrogen combined with hydrogen 30 g of ammonia is made ? Calculate the percentage yield ? N₂ + 3H₂                        2NH₃ N₂ = NH₃ 1mole = 2moles of ammonia 28g = 32g of ammonia Expected = 32g Observed = 30g % yield = 30/32 x 100 = 93.75% When 80 g of iron oxide reacts with carbon, 20 g of iron is produced. Calculate the % yield ? 2Fe₂O₃ + 3C               Fe + 3CO₂ Fe₂O₃ = Fe 2 moles = 4 moles 320g = 224 g 1g =224/320 80g = 224/320 x 80 = 56g 20/56 x 100 = 35.71% Banner 7

WHY PERCENTAGE YEILD IS NOT 100 %

The percentage yield can never by greater than 100%. It is very difficult to get 100% percentage yield but scientist always look for that route that gives maximum percentage yield.

• Reason for not getting 100% yield
• The reaction does not go to completion so complete products are not formed.
• Some of the reaction can start moving to the reverse direction if they are reversible
• Some of the reaction can go and form alterative or unwanted product.
• During the reaction some of the reactants can get lost or stick to the reaction vessel so do not react.
• If the reaction involves gaseous reactants they can escape.
• Some of the products can also get lost in the reaction vessel. If the reaction involves gaseous products they can also escape.
• The reagent might not be pure therefore did not react completely to give the desired yield.

Atom Economy

Atom economy = Mr of desired product/Total Mr of all the Products x 100 Calcium Oxide is produced using the following reaction CaCO₃                   CaO + CO₂ Calculate the atom economy Desired Product = CaO =56g Mass of reactants = CaCO₃ = 100g Atom Economy = 56/100 x 100 = 56% Iron oxide is reduced by carbon to form iron and carbon monoxide Fe₂0₃ + 3C                   2Fe +3CO Calculate the atom economy Desired Product = 2Fe =56 x 2 = 112g Mass of reactants = Mass of Fe₂CO₃ + 3(Mass of C) = 196g % Yield = 112/196 x 100 = 57%

REACTIONS WITH 100 % ATOM ECONOMY

• Addition reaction
• Reactions with only one product have 100 atom economy
• How to increase atom economy?
• Chemist should look for reaction that produce single product
• If the by products are produced they should look for recycling the byproducts or use them in some other reactions to increase the atom economy.

Titration

It is the technique used to determine the exact volume of acid and base to carry out a neutralization reaction.

TITRATION PROCEDURE

• Known Concentration solution either acid or base is measured by the pipette and is added into the conical flask. For this solution both concentration and volume is known.
• The indicator is also added in the conical flask. When the indicator changes colour the end point is reached i.e the solution gets completely neutralised.
• The unknown concentration solution is added into the burette. The starting volume is noted from the burette. The tap is then opened and the unknown solution is added dropwise into the conical flask with regular mixing.
• As soon as the indicator changes colour, the tap is closed and the final reading from the burette is noted.
• The entire process is repeated three times and the values are noted in the following format.
• The concordont reading are taken. The anamolous results are not taken into account.
• The means of the concordont readings are noted and used in the calculation

Titration Example

10 cm³ of 0.5 mol dm³ of sulphuric acid is titrated with sodium hydroxide. The results are given in the table. 2NaOH + H₂SO₄                  Na₂SO₄ + H₂0 Calculate the concentration of sodium hydroxide required to completely neutralize the acid ? Steps

1. Write the balanced chemical equation

2NaOH + H₂SO₄                  2NaOH + H₂0

1. Underneath each equation write the numerical value given for each

Moles = 5 x 10^-3 x 2 = 10^-2 moles

1. The quantity that has two value use the concentration triangle to find the moles

V = 10cm³ = 0.01dm³ C = 0.5 mol dm^-3 Moles = C x V 0.01 x 0.5=5 x 10^-3 moles

1. Use the molar ratio to find the moles of unknown quantity
2. Use titration volume and find the concentration

Mean Titre = 25.24+25.34+ 25.29/3 = 25.29cm3 Q1 In a Titration 50 cm³ of 0.1 mol dm³ of potasisum hydroxide is neutralized by 20 cm3 of sulphuric Acid. Calculate the concentration of sulphuric acid.

1. a) Write the balanced chemical equation

2KOH + H₂SO₄                        K₂SO₄ + 2H₂O

1. b) Underneath each equation write the numerical value given for each
2. c) The quantity that has two value use the concentration triangle to find the moles
3. d) Use the molar ratio to find the moles of unknown quantity
4. e) Use titration volume and find the concentration

V = 0.05dm³ V = 0.02 dm³ C = 0.1 mol dm^-3 = 0.005 moles m = 0.0025 mole C = 0.125 mol dm^-3 Key Terms

• Relative atomic mass — It is the ratio of the average mass of an atom compared to one twelfth of the mass of carbon-12
• Relative formula mass — It is the sum of relative atomic masses of all the atoms present in a formulae
• Moles – It is the amount of substance that has the same number of particles found in 12 g of carbon-12.
• Avogadros Constant —Number of particles present in one mole of the substance.
• 1mole = 6.02 x 10²³ atoms
• Limiting Reagent – It is the reagent that is completely used up in the reaction
• Yield— The mass of desired product obtained in a chemical reaction
• Percentage yield = Actual yeild /Theoretical Yield x 100
• Atom Economy = Mass of the desired product/total mass of all the reactants x 100

Concentration = Mass of Solute(g)/Volume of Solution(dm³) Titration — It is the technique used to determine the exact volume of acids and alkali required to carry out complete neutralization. Neutralization Reaction Reaction in which acid and base react to form salt and water. Acids—Substance that has pH less than T Alkali — Soluble bases that has pH greater than 7 Pipette – It is a glass tube with a bulge in the middle that is used to take out the exact volume of known concentration solutiom Burette— It is a long tube with the.tap at the bottom that is used to measure the titre Concordant – Values which are in the range of difference 0.1 to 0.2 cm³

TEST YOURSELF

Q1 Calculate the number of moles in the following

1. 10 g of calcium carbonate

Mr = CaCO₃ = 100g = 10/100 = 0.1 moles

1. 98 g of sulphuric acid

Mr = H₂SO₄ =98/98 = 1 mole

1. 18 g glucose

C₆H₁₂0₆ Mr = 180 18/180 = 0.1 moles

1. 90 cm3 of oxygen gas

Firstly convert 90cm³ in to dm³ 90/1000 = 0.09 dm³ 0.09/24 = 0.00375 moles

1. 10 cm3 of 0.2 mol dm3 sodium hydroxide solution

0.2 x 10/1000 = 0.002 moles   Q2 What is the mass of magnesium required to form 160 g of magnesium oxide ? MgO    =       Mg Mg + O₂                   2MgO MgO    =       Mg Moles     1                  1 Mass 4Og               24g 160g              96g = 96g Banner 8 Q3 Hydrogen react with chlorine to form hydrogen chloride gas. When 71 g of chlorine reacts 70 g of hydrogen chloride is obtained. Calculate the percentage yield. H₂ + Cl₂                             2HCl Cl₂                            HCl Cl₂    =       HCl Moles     1                  2 Mass   71g               73g 160g              96g % yield = 70/73 x 100 = 95.8% Q4 Calculate the atom economy of both of these reactions:- N₂ + 3H₂                             2NH₃ 34/28+6 x 100 = 100% CaCO₃                             CaO + CO₂ 40+16/100 x100 = 56% Q5 50 cm3 of 0.2 mol dm3 of HCI reacts with 10 cm3 of NaOH. Determine the concentration of NaOH ? NaOH + HCl                             NaCl + H₂O V = 50/1000 = 0.05dm³ C = 0.2 Moles = 0.1 NaOH V = 10/1000 = 0.01dm³ Moles = 0.1 moles C = 10 moldm^-3 Banner 9 Disclaimer: I have tried my level best to cover the maximum of your specification. But this is not the alternative to the textbook. You should cover the specification or the textbook thoroughly. This is the quick revision to help you cover the gist of everything. In case you spot any errors then do let us know and we will rectify it. References: BBC Bitesize Wikipedia Wikimedia Commons Image Source: Wikipedia Wikimedia Commons Flickr Pixabay