1. A 10.0 g sample of ethanol, C2H5OH, was boiled
with excess acetic acid, CH3COOH, to produce 14.8 g
of ethyl acetate, CH3COOC2H5. What is the
percent yield?

2. Ch3 #52
Sequential Reactions
CH3CH2Cl + Mg CH3CH2MgCl
CH3CH2MgCl + H2O CH3CH3 + Mg(OH)Cl
The Grignard reaction is a two-step reaction used to prepare pure
hydrocarbons. We allow 27.2g of CH3CH2Cl to react with excess
magnesium. CH3CH2MgCl is obtained in 79.5% yield. In the second
step a 85.7% yield of CH3CH3 is obtained. What mass of CH3CH3 is
collected?
1 mol CH3CH2Cl
64.4 g CH3CH2Cl
1 mol CH3CH2Cl
27.2g CH3CH2Cl * * = mol
mol CH3CH2MgCl*0.795= mol CH3CH2MgCl
1 mol CH3CH3
1 mol CH3CH2MgCl
0.3358mol CH3CH2MgCl* *0.857=0.2878mol CH3CH3
30.0g CH3CH3
1 mol CH3CH3
mol CH3CH3* = 8.63 g CH3CH3

3. Solutions
Parts of A (by mass)
100 parts mixture (by mass)
% Composition
%A (mass)=
grams of Solute
100 g of Solution
% Solute=
Mass of Solution = Mass of Solute + Mass of Solvent

4. What mass of NaOH is required to prepare 250.0 g
of solution that is 8.00% w/w NaOH?
8.00 g NaOH
100 g solution
* g solution = 20.0 g NaOH
Calculate the mass of 8.00% w/w NaOH solution that
contains 32.0 g of NaOH
100 g solution
8.00 g NaOH
* 32.0g NaOH= g Solution

5. Calculate the molarity of a solution that contains
moles of solute
L of solution
Molarity= = M
Calculate the molarity of a solution that contains
12.5 g of sulfuric acid in 1.75 L of solution.
12.5 g H2SO4
1.75 L solution
1 mole H2SO4
98.1 g H2SO4
* = M

6. Determine the mass of calcium nitrate required to
prepare 3.50 L of M Ca(NO3)2

7. The specific gravity of concentrated HCl is 1.185 and
it is 36.31% HCl. What is its molarity?
36.31 g HCl
100 g solution
Sp. Gr. =  1.185 g soln
ml of soln

8. Dilution M1V1=M2V2 What is the molarity of a solution that is
prepared with 2.00ml of 4.00*10-2M K2CrO4
and brought up to a volume of 500ml?
M1= 4.00*10-2M V1=2.00ml
M2=? V2=500ml
4.00*10-2M*2.00ml=XM*500ml
4.00*10-2M*2.00ml
500 ml
X= = 1.60*10-4M

9. What volume of 0.500 M BaCl2 is required to
completely react with 4.32 g of Na2SO4?
mol Na2SO4
142 g Na2SO4
mol BaCl2
mol Na2SO4
1 L solution
0.500 moles BaCl2
4.32 g Na2SO4 * * *
= L of 0.500M BaCl2 solution = 60.8ml of 0.500M BaCl2 solution