Need help with Pre Lab, Synthesis of alum PLEASE HELP Cesium Chromium alum was s

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.

Here is the chromium alum: KCr(SO4)2·12(H2O)

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 m

Explanation / Answer

1) Molecular weight of CsCr(SO4)2.12 H2O = 593 g/mol

2) Number of mmol of Chromium, Cr = (1.48 g/52 g/mol)*1000 = 28.5 mmol

3) Number of mmol of ceasium hydroxide = (1.18 mol/L)(20.7 mL) = 24.4 mmol

4) Number of mmol of H2SO4 = (9.36 mol/L)(10.44 mL) = 97.7 mmol

5) Ceasium hydroxide is the limiting reagent as it produces very less yield.

6) Number of mmol of water is consumed = (24.4 mmol)(12) = 292.8 mmol

7) Theoretical yield = (24.4 mmol)(593 g/mol) = 14.5 g

8) Percent yield = (6.23 g/14.5 g)*100 = 43 %

9) The solubility is 24.2 g/ 1000 mL, hence in 27.4 mL, the amount of alum left = (24.2 g/1000 mL)(27.4 mL) = 0.663 g

10)

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