Experiment 8: Limiting Reactant

Joseph Personelli

Pooja Patel

Dr. Mare Cudic

2/13/2014

Experiment 8 was conducted so that we could determine the limiting reactant in a

salt mixture. We also performed this experiment to determine the percent composition of

each substance in a salt mixture. We dissolved a salt in water, heated it to about 75c, and

then filtered it through filter paper with the help of a gravity filtering apparatus. The

limiting reactant for the unknown “Hootie and the Blowfish” was determined to be

K2C2O4*H2O. A precipitate formed when two drops 0.5M of K2C2O4 was added to the

supernatant. Because the precipitate formed, this indicated that Ca was in excess and that

C2O4 was the limiting reactant in the original salt mixture. To make sure that C2O4 was

truly the limiting reactant we added two drops of 0.5 M of CaCl2 and no precipitate

formed. Calculating the percent of each substance from the whole substance is another

way we were able to determine that K2C2O4 is the limiting reactant and that CaCl2 is the

excess reactant. The percent composition of the limiting reactant was lower than that of

the excess reactant. One possible error that could have occurred during the experiment is

when we were heating the salt mixture the heat went beyond 75C and it could have

possible denatured. Another error that could have affected our results is that the heating

oven was not set to the highest setting and our substance was still wet when it came time

to weigh it again.

Trial 1 Calculations:

1. 0.1324g (1 mol CaC2O4/128.12) = 0.00103

2. 1.039-0.1324= .9066

0.9066(1 mol K2C2O4/166.22) = 0.00545g

3. 0.1324g (1 mol CaC2O4/128.12g)(1 mol K2C2O4/1 mol CaC2O4)(184.24 g/ 1 mol

K2C2O4) = 0.190 g

4. .1324 g (1 mol CaC2O4/128.12)(1 mol CaCl2H2O/ 1mol CaC2O4)(147.02/ 1 mol

CaCl2H2O) = .152 g

5. (.190/1.039)X100= 18.28%

6. [(1.039-.190)/1.039] = 81.71%

7. .00545 X .152 = 8.284X10-4

8. .152 - 8.284X10-4 = .1511

9.

A. Precipitation of CaC2O4*H2O from the Salt Mixture Trial 1 Trial 21. Mass of beaker (g) 104.1994 107.92232. Mass of beaker and salt mixture (g) 105.2384 108.96373. Mass of salt mixture (g) 1.0390 1.04144. Mass of filter paper (g) 0.4638 0.45947. Mass of oven-dried CaC2O4 (g) 0.1324 0.0789

B. Determination of Limiting Reactant Formula1. Limiting Reactant in Salt Mixture K2C2O4*H2O2. Excess Reactant in Salt Mixture CaCl2*2H2O

Data Analysis Trial 1 Trial 2

1. Moles of CaC2O4 precipitated (mol) 0.00103 mol CaC2O4

0.0006158 mol CaC2O4

2. Moles of limiting reactant in salt mixture (mol) 0.00545 mol K2C2O4 0.00579 mol CaC2O4

Formula of limiting hydrate K2C2O4 K2C2O4

3. Mass of limiting reactant in salt mixture (g) 0.190 g K2C2O4 0.1135 g K2C2O4

Formula of limiting hydrate K2C2O4 K2C2O4

4. Mass of excess reactant in salt mixture (g) 0.152 g CaCl2 0.0905 g CaCl2

Formula of excess hydrate CaCl2*2H2O CaCl2*2H2O5. Percent limiting reactant in salt mixture (%) 18.28% 10.90%Formula of limiting hydrate K2C2O4 K2C2O4

6. Percent excess reactant in salt mixture (%) 81.71% 89.10%Formula of excess hydrate CaCl2*2H2O CaCl2*2H2O7. Mass of excess reactant that reacted (g) 8.284X10-4g reacted 5.24X10-4g reactedFormula of excess hydrate CaCl2 CaCl2

8. Mass of excess reactant, unreacted (g) 0.1511 g unreacted 0.0900 g unreacted