Ex 7

‫بسم ال الرحمن الرحيم‬ Name: ayedzahran Group: Day:

Experiment 7 "The limiting reaction" _________________________________________________________________ I. Observations and results: Part A Mass of filter paper Mass of filter paper and dried BaSO4 Mass of formed BaSO4 Moles of BaSO4 reacted Moles of Na2SO4 reacted Mass of Na2SO4 reacted Moles of Ba(NO3)2 reacted Mass of Ba(NO3)2 reacted

0.76 g 1.16 g 0.4 g (1) 1.7*10-3 mol (2) 1.7*10-3 mol (3) 0.24 g (4) 1.7*10-3 mol (3) 0.44 g (5)

Part B The limiting reactant is

Ba(NO3)2

Part C Mass of additional BaSO4 formed Moles of additional BaSO4 formed Moles of excess reactant remained Mass of excess reactant remained % limiting reactant in the mixture

1.03 - 0.77 = 0.26 g 1.1*10-3 mol (6) 1.1*10-3 mol (7) 0.29 g (8) 45.4 % (9)

II. Calculation: Na2SO4 + Ba(NO3)2 SO42- + Ba2+ (1) Mass BaSO4 = 1.16-0.76=0.4 g

BaSO4 + 2NaNO3 BaSO4

(2) Moles of BaSO4 =Mass/Molar mass =0.4/233.4=1.7*10-3 (3) From first equation

(4) Mass Na2SO4 = 1.7*10-3 *142=0.24 g (5) Mass Ba(NO3)2 = 1.7*10-3 *261.3 =0.44 g (6) Moles of additional BaSO4 = 0.26/233.4 = 1.1*10-3 mol

(7) From second equation (8) Mass excess Ba(NO3)2 = 1.1*10-3 * 261.3 = 0.29 g (9) % Ba(NO3)2 =mass Ba(NO3)2 *100%/(mass Ba(NO3)2 + mass Na2SO4

excess + mass Na2SO4 reactant ) = 0.44 * 100% / (0.44+0.24+0.29) = 45.4% III. Answer the following questions: 1. Name the advantages of using an excess of one reactant in performing chemical reactions Usually the less expensive and the more abundant is excess and the advantages is speed the rat of reaction and increase the percentage yield of the production 2. How would your experiment results be effected if : a. The separated precipitate was not dried enough before weighing The weight of precipitate will increase, the moles of precipitate will increase, the moles of limiting reactant will increase, the mass of limiting reactant will increase, the moles of excess reactant will increase and the mass of excess reactant will increase, so the actual numbers will increase and be wrong b. Insufficient amount of the precipitating agent was added in part B The amount of excess will be less; the percentage of limiting reactant will increase in the reaction 3. If a 1.5 g of sodium carbonate – calcium chloride was dissolved in water

and 0.63 g of solid calcium carbonate was filtered and dried, calculate the percentage composition of the mixture if the limiting reactant was determined as sodium carbonate Na2CO3 +CaCl2 CaCO3+2NaCl -3 Mol CaCO3 = 0.63/100 = 6.3*10 mol Mol Na2CO3 = Mol CaCO3 =6.3*10-3 mol Mass Na2CO3 = 6.3*10-3 * 106 = 0.6678 g % Na2CO3 = 0.6678 *100%\1.5 = 44.52% %CaCl2 = 100% - %Na2CO3 = 100% - 44.52% = 55.48% Not: For experiment number , For calculation number , For the solution of answers .