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Experiment 7
"The limiting reaction"
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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
Mol CaCO3 = 0.63/100 = 6.3*10 ³ 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 .