Stoichiometry Lab Report
Stoichiometry lab 1
Purpose: The purpose of this lab is to find the limiting reactant, also to find the percentage yield and percentage purity of the reaction that happens between Calcium Chloride and Sodium Carbonate. The other purpose was to know how the reaction can be balanced and created.
Hypothesis: In this lab we are going to see a precipitation reaction. This is a reaction where two soluble salts Sodium Carbonate and Calcium Chloride are added together and the result is the precipitation of single Product while the other product remains in solution. This means by using the moles in this lab we could find the limiting reactant, percentage yield and the percent purity
Materials:
* Electronic Scale * Sodium Carbonate * Calcium Chloride * Wayne Paper …show more content…
* Watch Glass * Funnel * Two 100 mL Beakers
Safety equipment: - * The hair should be tied if it’s long. * Safety Goggles. * No jackets or long clothes. * Work only with experiment materials.
Procedure:
1. Take two clean, dry, 100 mL beakers. Label the 1st beaker "Na2CO3", & the 2nd "CaCl2".
2. Add your mass of Na2CO3 to the 1st beaker . (Record mass of Na2CO3)
3. Add your mass of CaCl2 to the 2nd beaker. (Record mass of CaCl2)
4. Dissolve each of the solids in approximately 25 mL of de-ionized water. (Rate of dissolution can be increased by agitation)
5. Note the color of the two solutions (Record color)
6. Add a few millilitres of the CaCl2 solution to the beaker containing Na2CO3.
7. Slowly add the remaining CaCl2 solution to the Na2CO3, a few millilitres at a time.
8. Rinse remaining traces of the CaCl2 solution into the flask with approximately 5 mL of deionized water.
Repeat a second time to ensure all CaCl2 is transferred to the beaker.
9.
Stir the reaction mixture for a few minutes.
10. Obtain a piece of filter paper, determine its mass. (Record mass)
11. Set up the apparatus (funnel, ring stand, filter paper) for filtration while the precipitate in the beaker settles.
12. Place the empty beaker under the funnel and transfer the reaction mixture to the filter paper. (Careful! Do not allow reaction mixture to rise above the rim of the filter paper)
13. Add approximately 10 mL of distilled water to the reaction mixture remaining in the beaker. Transfer the diluted reaction mixture to the filter paper. 14. Repeat step #13. (Observe the color of the filtrate)
15. Label a watch-glass with: your name, date, and CaCO3 .
16. When the filtration process is complete, place the filter paper and the solid CaCO3 product on the labeled watch-glass.
17. Leave your labeled watch glass out for drying overnight (low temperature).
18. Record the mass of the dry filter paper + CaCO3. Record mass
Questions and answers for Limiting Reactant
1. The mole of Na2CO3 used is n= m÷M
= 3.01g÷105.99 =0.028mol
2. The moles of CaCO3 n= m÷M
= 4g÷100.09g/mol =0.040 mol
3a: mass of CaCO3 produced is m=n×M = 0.028mol×100.09g/mol
=2.80g
b. Mole ratio: Na2CO3: CaCO3 1: 1
4. The actual number of moles reacted is 0.028 mol like it is the limiting agent.
5: The Limiting reactant is = Na2CO3 – number of moles left = 0.036 - 0.028
= 0.008 mol m= 0.008 ×105.99
= 0.848g
∴ The Sodium Carbonate was in excess of 0.848g.
6: What was the percentage yield of the reaction?
Precipitate Mass = total mass - filter paper mass
=4.35 - 1.60 =2.75g
∴ Percentage yield= (2.75g÷2.80g) ×100%
=98.2%
7. Percent Purity= m of precipitate/m of reactant percentage Na2CO3 = (2.75÷3) × 100% = 91.7% percentage CaCl2= (2.75÷4) × 100% = 68.75%
Answers for the Non- Limiting Reactant
1. The moles of Na2CO3 used (form table masses).
Number of moles = 3.18÷105.99 = 0.036mol
2. The moles of CaCO3 produced. n= 4 ÷110.98 = 0.036mol
3a: The mass of CaCO3 produced is m= 0.036×100.09=3.60g
B. Calculate moles of CaCO3 Mole ratio is = 1:1
∴ Number of moles =0.036mol
4: In consideration of the chemical equation and the moles of CaCO3 calculated in step 3B, determine the actual number of moles of Na2CO3 and CaCl2 that reacted.
The two reactants has an equal amount of moles of 0.036.
5. There was no limiting agent because both of the reactants were equal. Therefore there is also no excess reactant.
6. The mass of the precipitate = total mass - filter paper mass
=2.83g - 0.88g =1.95 g
Percent yield= (1.95 ÷ 3.60) × 100% = 54.2%
7. Percent purity= mass of precipitate ÷ mass of reactant
Percent purity of Na2CO3 = 1.95 ÷3.9 = 50%
Percent purity of CaCl2= 1.95 ÷ 4.1 = 47.6%
Purpose: The purpose of this lab is to find the limiting reactant, also to find the percentage yield and percentage purity of the reaction that happens between Calcium Chloride and Sodium Carbonate. The other purpose was to know how the reaction can be balanced and created.
Hypothesis: In this lab we are going to see a precipitation reaction. This is a reaction where two soluble salts Sodium Carbonate and Calcium Chloride are added together and the result is the precipitation of single Product while the other product remains in solution. This means by using the moles in this lab we could find the limiting reactant, percentage yield and the percent purity
Materials:
* Electronic Scale * Sodium Carbonate * Calcium Chloride * Wayne Paper …show more content…
* Watch Glass * Funnel * Two 100 mL Beakers
Safety equipment: - * The hair should be tied if it’s long. * Safety Goggles. * No jackets or long clothes. * Work only with experiment materials.
Procedure:
1. Take two clean, dry, 100 mL beakers. Label the 1st beaker "Na2CO3", & the 2nd "CaCl2".
2. Add your mass of Na2CO3 to the 1st beaker . (Record mass of Na2CO3)
3. Add your mass of CaCl2 to the 2nd beaker. (Record mass of CaCl2)
4. Dissolve each of the solids in approximately 25 mL of de-ionized water. (Rate of dissolution can be increased by agitation)
5. Note the color of the two solutions (Record color)
6. Add a few millilitres of the CaCl2 solution to the beaker containing Na2CO3.
7. Slowly add the remaining CaCl2 solution to the Na2CO3, a few millilitres at a time.
8. Rinse remaining traces of the CaCl2 solution into the flask with approximately 5 mL of deionized water.
Repeat a second time to ensure all CaCl2 is transferred to the beaker.
9.
Stir the reaction mixture for a few minutes.
10. Obtain a piece of filter paper, determine its mass. (Record mass)
11. Set up the apparatus (funnel, ring stand, filter paper) for filtration while the precipitate in the beaker settles.
12. Place the empty beaker under the funnel and transfer the reaction mixture to the filter paper. (Careful! Do not allow reaction mixture to rise above the rim of the filter paper)
13. Add approximately 10 mL of distilled water to the reaction mixture remaining in the beaker. Transfer the diluted reaction mixture to the filter paper. 14. Repeat step #13. (Observe the color of the filtrate)
15. Label a watch-glass with: your name, date, and CaCO3 .
16. When the filtration process is complete, place the filter paper and the solid CaCO3 product on the labeled watch-glass.
17. Leave your labeled watch glass out for drying overnight (low temperature).
18. Record the mass of the dry filter paper + CaCO3. Record mass
Questions and answers for Limiting Reactant
1. The mole of Na2CO3 used is n= m÷M
= 3.01g÷105.99 =0.028mol
2. The moles of CaCO3 n= m÷M
= 4g÷100.09g/mol =0.040 mol
3a: mass of CaCO3 produced is m=n×M = 0.028mol×100.09g/mol
=2.80g
b. Mole ratio: Na2CO3: CaCO3 1: 1
4. The actual number of moles reacted is 0.028 mol like it is the limiting agent.
5: The Limiting reactant is = Na2CO3 – number of moles left = 0.036 - 0.028
= 0.008 mol m= 0.008 ×105.99
= 0.848g
∴ The Sodium Carbonate was in excess of 0.848g.
6: What was the percentage yield of the reaction?
Precipitate Mass = total mass - filter paper mass
=4.35 - 1.60 =2.75g
∴ Percentage yield= (2.75g÷2.80g) ×100%
=98.2%
7. Percent Purity= m of precipitate/m of reactant percentage Na2CO3 = (2.75÷3) × 100% = 91.7% percentage CaCl2= (2.75÷4) × 100% = 68.75%
Answers for the Non- Limiting Reactant
1. The moles of Na2CO3 used (form table masses).
Number of moles = 3.18÷105.99 = 0.036mol
2. The moles of CaCO3 produced. n= 4 ÷110.98 = 0.036mol
3a: The mass of CaCO3 produced is m= 0.036×100.09=3.60g
B. Calculate moles of CaCO3 Mole ratio is = 1:1
∴ Number of moles =0.036mol
4: In consideration of the chemical equation and the moles of CaCO3 calculated in step 3B, determine the actual number of moles of Na2CO3 and CaCl2 that reacted.
The two reactants has an equal amount of moles of 0.036.
5. There was no limiting agent because both of the reactants were equal. Therefore there is also no excess reactant.
6. The mass of the precipitate = total mass - filter paper mass
=2.83g - 0.88g =1.95 g
Percent yield= (1.95 ÷ 3.60) × 100% = 54.2%
7. Percent purity= mass of precipitate ÷ mass of reactant
Percent purity of Na2CO3 = 1.95 ÷3.9 = 50%
Percent purity of CaCl2= 1.95 ÷ 4.1 = 47.6%