Need help with Pre Lab, Synthesis of alum PLEASE HELP Cesium Chromium alum was s
ID: 1004953 • Letter: N
Question
Need help with Pre Lab, Synthesis of alum PLEASE HELP
Cesium Chromium alum was synthesized according to the procedure in SUSB-030. The following table shows the reagents used in the synthesis.
Question 1
What is the molar mass of Cesium Chromium Alum? (use the atomic masses in the table at the back of the lab manual. Give the result to 3 significant figures.)
Enter Your Answer: 474 Incorrect
Question 2
How many mmol of Chromium were used in the synthesis?
Enter Your Answer: .028 Incorrect
Question 3
How many mmol of Cesium hydroxide were used in the synthesis?
Enter Your Answer: 1.18 Incorrect
Question 4
How many mmol of sulfuric acid were used in the synthesis?
Enter Your Answer: 9.36 Incorrect
Question 5
What was the limiting reagent?
A. Chromium B. Cesium Hydroxide C. Sulfuric Acid D. Water
E. The mmol of Chromium and Cesium Hydroxide are equal, so either of them is limiting
Enter Your Answer: A B C D E F E Incorrect
Question 6
How many mmols of water are consumed in making the product?
Enter Your Answer:
Question 7
What is the theoretical yield of alum in grams?
Enter Your Answer: 110 Incorrect
After reaction, the solution is evaporated to reduce its volume.
The final volume is 27.4 mL. The solution is cooled in an ice bath.
6.23 g of alum is recovered by filtration.
Question 8
What is the percent yield?
Enter Your Answer: 5.7% Incorrect
Question 9
The solubility of the alum at 0 oC is 24.2 g/L. How many grams of alum is left in solution after evaporation?
Enter Your Answer:
Question 10
What percent of the actual yield does the alum in solution represent?
Enter Your Answer: .2 Incorrect
Value Units Weight of Chromium 1.48 g Concentration of Cesium Hydroxide 1.18 mol/L Volume of Cesium Hydroxide 20.7 mL Concentration of Sulfuric Acid 9.36 mol/L Volume of Sulfuric Acid 10.44 mExplanation / Answer
1. Alum has formula of KAl(SO4)2- ·12H2O.
In our case K is replaced by Chromium and Cesium making it CsCr(SO4)2- ·12H2O
Molar mass = 593 g/mol
2.
Your answer is correct but units are wrong. Units asked are in mmol.
Answer is 28.46 mmol
3.
Concentration of CsOH was 1.18 mol/L but we have used 20.7 ml only.
mMoles in 20.7 ml = 1.18/1000 * 20.7 = 0.0244 mol = 24.4 mmol
4.
Concentration of H2SO4 was 9.36 mol/L and we have used 10.44 ml only.
mMoles in 10.44 ml = 9.36/1000 * 10.44 = 0.0977 mol = 97.7 mmol
5.
Limiting reagent is the one whose concentration is lowest thereby limiting the reaction to its own concentration.
Above calculations show that CsOH has the lowest concentration as 24.4 mmol. Therefore, CsOH is limiting reagent.
B. Cesium Hydroxide
6. mmol of water
We don’t have volume. Do you have any more data?
Just divide the ml with molar mass of water (18). Then convert moles to mmol.
7.
We will require to convert mmol to grams
Mmol of each will be reduced to 24.4 mmol as it is concentration of limiting reagent.
Grams of CsOH = 24.4 mmol = 3.65 g
Grams of H2SO4 = 24.4 mmol = 2.39 g
Grams of Chromium = 24.4 mmol = 1.26 g
CsOH + H2SO4 + Cr = Weight of Alum
3.65 + 2.39 + 1.26 = 7.3 g
Theoretical yield is 7.3 g
8.
% yield = Actual yield /theoretical yield * 100 = 6.23 / 7.3 *100
= 85.34 %
9.
Solubility is 24.2 g/L
But we had 27.4 mL solution. According to solubility, highest amount soluble is
= 24.2/1000 * 27.4 = 0.6 g
10.
Alum in solution represent = 0.6/6.23 *100 = 9.63%