CHM 2045C    Module Five Exam-page 6

Module Four-Part K Excess-Limiting Reagent Problem         4 points

How many grams of Calcium phosphate can be made according to the reaction (unbalanced):

3 CaCl2       +    2 K3PO4     ---->        Ca3(PO4)2        +  6 KCl

by mixing a solution of 5.00 grams of CaCl2   with another containing 8.00 grams of Potassium phosphate?

!st Balance the equation: 2nd: Calculate the Molecular Masses of three of the salts:   Use Dimensional Analysis to calculate the maximum amount of Calcium Phosphate that can be produced from 5.00 g of Calcium Chloride:

____?____g Ca3(PO4)2   =  5.00 g CaCl2 4.658 g Ca3(PO4)2   could be produced if all the 5.00 grams of calcium Chloride is used up in the reaction.

Next calculate the maximum amount of Calcium Phosphate that can be produced from reacting all the 8.00 g of Potassium Phosphate:

____?____g Ca3(PO4)2   =  8.00 g K3 PO4 5.845 g Ca3(PO4)2   could be produced if all the 8.00 grams of Potassium Phosphate is used up in the reaction.

Therefore since less is produced from the 5.00 g of CaCl2 , 4.658 g Ca3(PO4)2    is the answer

The answer can be obtained in seconds using Chem-i-Calc:

1st substitute 5.00 grams of CaCl2 in Reaction Mode of Chem-i-Calc: 4.658 g Ca3(PO4)2   is produced

2nd substitute 8.00 g of K3PO4 in Chem-i-Calc: 5.845 g Ca3(PO4)2   is produced

Therefore 4.658 g Ca3(PO4)2   is the answer

Module Four-Part L   Impure Reagents/% Yield Stoichiometry Problem 4 points

A laboratory manual calls for 13.0 grams of butanol,  21.6 grams of sodium bromide, and 33.8 grams of sulfuric acid as reactants in this reaction:

C4H9OH   +   NaBr     +     H2SO4   ------>     C4H9Br   +   NaHSO4    +   H2O

A student following these directions obtains 16.8 grams of butyl bromide (C4H9Br).   What are the theoretical yield and percent yield of this reaction?

1st Balance the Equation:

C4H9OH   +   NaBr     +     H2SO4   ------>     C4H9Br   +   NaHSO4    +   H2O

It is balanced as all the coefficients are one (1).

2nd Calculate the molar masses of reactants and the major product:

C4H9OH:      74.12g C4H9OH = 1 mol C4H9OH

NaBr:         102.89 g NaBr  = 1 mol NaBr

H2SO4           98.08 g H2SO4 = 1 mol H2SO4

C4H9Br       137.02 g C4H9Br = 1 mol C4H9Br

Find:

C4H9Br   Theoretical yield of C4H9Br : ___?___

1st Check 13.0 grams of butanol (which in most organic reactions is the limit)

13.0g  C4H9OH = _24.0_g C4H9Br 2nd Check the 21.6 grams of Sodium Bromide:

21.6g  NaBr = _28.8_g C4H9Br 3rd Check the 33.8 grams of Sulfuric Acid

33.8 g  H2SO4  = _47.2_g C4H9Br My assumption is correct, the Butanol is the limiting reagent:

Per Cent Yield:

Actual/Theoretical x 100  = % yield

16.8 g actual C4H9Br        100

----------------------------- X  --------  =  70.0 % yield

24.0 g theorl C4H9Br         100