Objective:
1. To study the quantitative relationship between the amount of reactants and products of a reaction.
2. To investigate the reaction of magnesium with hydrochloric acid
3. To identify the unknown X value in the chemical equation between magnesium and hydrochloric acid
4. To determine the valency of magnesium
Introduction:
Stoichiometry is the study of the quantitative relationship between amounts of reactants and products of a reaction. Stoichiometry canbe used to calculate the amounts of products with given the reactants and percent yield. In this experiment, a known starting mass of magnesium and the measured collection of hydrogen gas will be used to determine the reaction stoichiometry.
Magnesium is used to reacts with hydrochloric acid in order to produce hydrogen gases. One of the purpose of the experiments is to determine the value of X in the following equation:
Mg (s) + X HCl (aq) à MaCl2(aq) + X/2 H2(g)
The mass of magnesium is measured by the analytical balance before used in the experiment. A known amount of magnesium is reacted with a large amount of excess hydrochloric acid and hence the magnesium acts as a limiting agent in the particular reaction. As the amount of HCl is used in excess, this can ensure that all the magnesium will be reacted completely in the reaction. The formation of MgClx and H2 are depends on the amount of magnesium used. The comparison of the amount of hydrogen gas produced with the amount of magnesium consumed will enable the X value to be determined.
Materials: magnesium ribbon, 0.5M hydrochloric acid, cotton, distilled water
Apparatus: 50ml burette, 25ml pipette, retort stand, analytical balance, watch glass, beaker, gauze, funnel, glass rod, thermometer
Procedure:
1. Upside down the burette and determine the volume of unmarked space in a clean and dry 50ml burette by pipetting 25ml of water into the vertically clamped burette.
2. Note the burette reading, drain the burette and repeat. Leave the water in the buretee for 10 minutes and check whether leaks occur.
3. A piece of magnesium ribbon is cleaned with steel wool. A piece of magnesium is cut off by scissors to a smaller size. Curl up the ribbon. Tare a watch glass on the four decimal balance and accurately weigh between 0.03g and 0.036g of the magnesium on the watch glass.
4. A small filter funnel with short stem (1cm-1.5cm) is covered with gauze and then is inverted. The filter funnel with gauze is placed on the watch glass over the magnesium.
5. The beaker is filled with tap distilled water until the level is approximately above the end of the funnel stem by 0.5cm-1.0cm. The burette is completely filled with 0.5M HCl and invert it(with supplied cork pressed flat over the open end) and place it in the water in the beaker.
6. Remove the cork and place the end of the burette over the stem of the funnel, ensuring that no air enters and clamp it into position.
7. Remove the excess of water with a pipette until the level is just above the stem if the funnel.
8. Add about 100ml of 0.5M of HCl to the beaker, use a glass rod to stir the water to ensure complete mixing such that the HCl reachest the magnesium. This may be helped by tapping the watch glass gently with a glass rod.
9. Stir the solution to initiate the reaction and then do not stir further so that the reaction proceeds unaided. At the completion of the reaction, tap the watch glas gently to dislodge any gas bubbles.
Result and calculation:
Mass of magnesium ribbon= 0.0356g
Initial reading of burette = 5cm3
Final reading of burette = 37cm3
Volume of unmarked place = 5cm3
Total volume of H2 collected = 35cm3
Mg (s) + X HCl (aq) à MaCl2(aq) + X/2 H2(g)
Mole of magnesium = 0.0356g/24 g mol-1
= 1.483x10-3 mol
Actual mole of HCl used = 100cm3 x 0.5M / 1000
= 0.05 mol
Number of mole of HCl / number of mole of Mg
= 0.05 mol/ 1.483x10-3 mol
= 33.72 > 1
Magnesium is a limiting agent.
Theoretical mole of hydrogen gas = 0.035dm3 / 24dm3 mol-1
= 1.483x10-3 mol
Number of moles of magnesium : number of mole of hydrogen gas
1.483x10-3 mol : 1.483x10-3 mol
1 : 1
Thus, X/2 = 1
X = 2
Discussion:
Magnesium is an alkaline earth metal which is categorized to the Group 2 element in the periodic table. All the alkaline metal reacts with acid will produce hydrogen gas. Since the HCl is added in excess, hence the magnesium is the limiting reactant in this reaction. Limiting reactant is the reactant which restricts the amount of products generated in the reaction. In this situation, the magnesium is used up completely to react with the excess hydrochloric acid.
From the calculation, the value of X in the chemical equation between magnesium and hydrochloric acid was determined. The value of X is 2. So that, the complete chemical reaction for magnesium and acid is Mg (s) + 2 HCl (aq) à MaCl2(aq) + H2(g). Based on the Las of Conservation of Matter, matter can be neither created or destroyed. An equation must have the same number of atoms of the same kind on both sides of the equation. Hence, 1.483x10-3 mol of magnesium was used up completely to react with 2.966x10-3 mol of acid to produce 1.483x10-3 mol of magnesium chloride and hydrogen gas.
The reaction between magnesium and hydrochloric acid is an exothermic reaction. This reaction releases heat energy from the reaction to its surrounding which caused the solution is the beaker become hotter. The temperature cannot be detected significantly because the rise in temperature was just slight different. The temperature is only increased by few degree Celsius due to the large specific heat capacity of the water.
The magnesium ribbon used in the experiment was in the curved-shape. The curved shape magnesium ribbon was used instead of the flat magnesium ribbon because the former one can reacts faster than the latter one. The curved magnesium own larger surface area which increase the reaction proceed faster compared to the flat magnesium ribbon although they are in the same volume. As the surface area is increased, the magnesium atoms can have more surfaces to be exposed to the hydrochloric acid. Thus, the reactivity of the reaction increases.
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ReplyDeleteHow did u calculate the volume of unmarked space?
ReplyDeleteAnd are the readings read upright or inverted? thks