JK-TK/Newdata huggingface dataset:
Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples: 1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1.
Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p) Chemical equation CaCO3(s) -> CaO(s) + CO2(g) Mole ratios 1: 1: 1 Molar Mass CaCO3 =100g Method 1 100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p 5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3 100g Method 2
19 19 Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles 100 g Mole ratio 1:1 Moles of CO2(g) = 0.05moles Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3 2. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen.
1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2.
If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 ) Chemical equation Copper does not react with hydrochloric acid 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g) Method 1 3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al 840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium 3 x 22.4 x 1000 Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g =0.325g of copper % copper = mass of copper x100% = 32.5% Mass of alloy Method 2 Mole ratio 2Al: 3H2 = 2:3 Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles Molar gas volume 22400cm3 Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g Total mass of alloy – mass of aluminium = mass of copper => 1.0g - 0.675g = 0.325 g of copper % copper = mass of copper x100% = 32.5% Mass of alloy
20 20 (f)Gay Lussac’s law Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure” Gay Lussacs law thus only apply to gases Given the volume of one gas reactant, the other gaseous reactants can be deduced thus: Examples 1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.
Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen. Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l) Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 2. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air) Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
21 21 Volume ratios 2 : 1 : 0 Reacting volumes 50cm3 : 25cm3 50cm3 of Oxygen is used 21% = 25cm3 100% = 100 x 25 = 21 3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy. Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.
Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g) Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3 Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5 Reacting volume ratios 1volume 3 volume 2 volume 2 volume Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g) If “4H” are in 2H2O(g) the y=4 If “2C” are in 2CO2 (g) the x=2 Thus(i) chemical formula of hydrocarbon = C2H4 (ii) chemical name of hydrocarbon = Ethene 4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.
Obtain the equation for the reaction and name the gaseous product. Chemical equation: NO (g) + O2 (g) -> NOx Volumes 100cm3 : 50cm3 : 100 Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50 Reacting volume ratios 2volume 1 volume 2 volume Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g) Thus(i) chemical formula of the nitrogen compound = 2 NO2 (ii) chemical name of compound = Nitrogen(IV)oxide 5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. 22 22 When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3. (a)What volume of Oxygen was used during the reaction.(1mk) Volume of Oxygen used =100-25 =75cm3√ (P was completely burnt) (b)Determine the molecular formula of the hydrocarbon(2mk) CxHy + O2 -> xCO2 + yH2O 15cm3 : 75cm3 15 15 1 : 3√ => 1 atom of C react with 6 (3x2)atoms of Oxygen Thus x = 1 and y = 2 => P has molecula formula CH4√ (g) Ionic equations An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products. Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions) An ionic equation is usually derived from a stoichiometric equation by using the following guidelines Guidelines for writing ionic equations 1.Write the balanced stoichiometric equation 2.Indicate the state symbols of the reactants and products 3.Split into cations and anions all the reactants and products that exist in aqueous state. 4.Cancel out any cation and anion that appear on both the product and reactant side. 5. Rewrite the chemical equation. It is an ionic equation.
5. Rewrite the chemical equation. It is an ionic equation. Practice (a)Precipitation of an insoluble salt
23 23 All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt 1. When silver nitrate(V) solution is added to sodium chloride solution,sodium nitrate(V) solution and a white precipitate of silver chloride are formed. Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.
Balanced stoichiometric equation AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq) Split reactants product existing in aqueous state as cation/anion Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Cancel out ions appearing on reactant and product side Ag+(aq) + NO3- (aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq)+ NO3- (aq) Rewrite the equation Ag+(aq) + Cl-(aq) -> AgCl(s) (ionic equation) 2. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.
When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed. Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known.
Balanced stoichiometric equation Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq) Split reactants product existing in aqueous state as cation/anion Ba2+(aq) + 2NO3- (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3- (aq)+ Cu2+(aq) Cancel out ions appearing on reactant and product side Ba2+(aq) + 2NO3- (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3- (aq) + Cu2+(aq) Rewrite the equation Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation) 3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide. Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.
Balanced stoichiometric equation Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
24 24 Split reactants product existing in aqueous state as cation/anion Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Cancel out ions appearing on reactant and product side Pb2+(aq) + 2NO3- (aq) + 2K +(aq) + 2I - (aq) -> PbI2 (s) + 2NO3- (aq)+ 2K +(aq) Rewrite the equation Pb2+(aq) + 2I - (aq) -> PbI2 (s) (ionic equation) (b)Neutralization Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base. (i)Reaction of alkalis with acids 1.Reaction of nitric(V)acid with potassium hydroxide Balanced stoichiometric equation HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq) Split reactants product existing in aqueous state as cation/anion H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Cancel out ions appearing on reactant and product side H+(aq) + NO3- (aq) + K +(aq) + OH - (aq) -> H2O (l) + NO3- (aq)+ K +(aq) Rewrite the equation H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 2.Reaction of sulphuric(VI)acid with ammonia solution Balanced stoichiometric equation H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH - (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
25 25 Rewrite the equation 2H+ (aq) + 2OH - (aq) -> 2H2O (l) H+ (aq) + OH - (aq) -> H2O (l) (ionic equation) 3.Reaction of hydrochloric acid with Zinc hydroxide Balanced stoichiometric equation 2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq) Split reactants product existing in aqueous state as cation/anion 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Cancel out ions appearing on reactant and product side 2H+(aq) + 2Cl- (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl- (aq)+ Zn 2+ (aq) Rewrite the equation 2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation) (h)Molar solutions A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M. Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.