JK-TK/Newdata huggingface dataset:
Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
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20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined.
41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
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20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined. e.g rusting ,decomposition of hydrogen peroxide and weathering. Some reactions are too fast and instantaneous e.g. neutralization of acid and bases/alkalis in aqueous solution and double decomposition/precipitation. Other reactions are explosive and very risky to carry out safely e.g. reaction of potassium with water and sodium with dilute acids. The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries. Theories of rates of reaction. The rate of a chemical reaction is defined as the rate of change of concentration/amount of reactants in unit time. It is also the rate of formation of given concentration of products in unit time. i.e. Rate of reaction = Change in concentration/amount of reactants Time taken for the change to occur Rate of reaction = Change in concentration/amount of products formed Time taken for the products to form For the above, therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product. The SI unit of time is second(s) but minutes and hours are also used. (a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction.
(a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction. Collisions that initiate a chemical reaction are called successful / fruitful/ effective collisions (iii)the speed at which particles collide is called collision frequency. The higher the collision frequency the higher the chances of successful / fruitful/ effective collisions to form products. (iv)the higher the chances of successful collisions, the faster the reaction. (v)the average distance between solid particles from one another is too big for them to meet and collide successfully. (vi)dissolving substances in a solvent ,make the solvent a medium for the reaction to take place. The solute particle distance is reduced as the particle ions are free to move in the solvent medium. (vii)successful collisions take place if the particles colliding have the required energy and right orientation which increases their vibration and intensity of successful / fruitful/ effective collisions to form products. (b)The Activation Energy(Ea) theory
The Enthalpy of activation(∆Ha) /Activation Energy(Ea) is the minimum amount of energy which the reactants must overcome before they react. Activation Energy(Ea) is usually required /needed in bond breaking of the reacting particles. Bond breaking is an endothermic process that require an energy input. The higher the bond energy the slower the reaction to start of. Activation energy does not influence whether a reaction is exothermic or endothermic. The energy level diagrams below shows the activation energy for exothermic and endothermic processes/reactions. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy.
Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy. Endothermic reaction cannot proceed without further heating /external energy because it does not generates its own energy/heat to overcome activation energy. It generally therefore requires continuous supply of more energy/heat to sustain it to completion. 3. Measuring the rate of a chemical reaction. The rate of a chemical reaction can be measure as: (i)Volume of a gas in unit time; - if reaction is producing a gas as one of the products. - if reaction is using a gas as one reactants (ii)Change in mass of reactants/products for solid products/reactants in unit time. (iii)formation of a given mass of precipitate in unit time (iv)a certain mass of reactants to completely form products/diminish. Reactants may be homogenous or heterogenous. -Homogenous reactions involve reactants in the same phase/state e.g. solid-solid,gas-gas,liquid-liquid. -Heterogenous reactions involve reactants in the different phase/state e.g. solid-liquid,gas-liquid,solid-gas. 4. Factors influencing/altering/affecting/determining rate of reaction The following factors alter/influence/affect/determine the rate of a chemical reaction: (a)Concentration (b)Pressure (c) Temperature (d)Surface area
(e)Catalyst a) Influence of concentration on rate of reaction The higher the concentration, the higher the rate of a chemical reaction. An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker.
An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock.
Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock. (ii)Consistent use of a minimum of four decimal points on inverse/reciprocal of time as evidence of understanding/knowledge of the degree of accuracy of scientific calculator. (c) accuracy against a school value based on candidate’s teachers-results submitted. (d) correct trend (time increase as more water is added/acid is diluted) in conformity with expected theoretical results. Sample questions 1. On separate graph papers plot a graph of: (i)volume of acid used(x-axis) against time. Label this graph I (ii) volume of acid used(x-axis) against 1/t. Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.
Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4.
Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker.
Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.
Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate.
Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1.
Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water (cm3) Volume of sodium thiosulphate (cm3) Concentation of sodium thisulphate in molesdm-3 Time(T) taken for mark ‘X’ to be invisible/ obscured(seconds) T-1 5.0 0.0 25.0 0.4 20.0 5.0 x 10-2 5.0 5.0 20.0 0.32 23.0 4.35 x 10-2 5.0 10.0 15.0 0.24 27.0 3.7 x 10-2 5.0 15.0 10.0 0.16 32.0 3.13 x 10-2 Note concentration of diluted solution is got: C1V1=C2V2 => 0.4 x 25 = C2x 25 =0.4M C1V1=C2V2 => 0.4 x 20 = C2x 25 =0.32M C1V1=C2V2 => 0.4 x 15 = C2x 25 =0.24M C1V1=C2V2 => 0.4 x 10 = C2x 25 =0.16M Sample questions 1. On separate graph papers plot a graph of: (i)Concentration of sodium thiosulphate against time. Label this graph I
(ii)Concentration of sodium thiosulphate against against T-1.Label this graph II 2. Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water.
Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M From correct graph at concentration 0.192M => 2.4 x10-2 I/t = 2.4 x10-2 t = 41.6667seconds (ii)22cm3 of sodium thisulphate is diluted with 3cm3 of water. At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used.
At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M Mole ratio Na2S2 O3 :HCl =1:2 Moles of Na2S2 O3 = 0.192M x 12 => 2.304 x 10-3 moles 1000 Mole ratio HCl =2.304 x 10-1 moles = 1.152 x 10-3 moles 2 Molarity o f HCl = 1.152 x 10-3 moles x 1000 = 0.2304M 5.0
Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 10centimeter length of magnesium ribbon with sand paper/steel wool. Measure 40cm3 of 0.5M dilute hydrochloric acid into a flask .Fill a graduated gas jar with water and invert it into a trough. Stopper the flask and set up the apparatus to collect the gas produced as in the set up below: Carefully remove the stopper, carefully put the magnesium ribbon into the flask . cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar.