Unit 3 Stoichiometry IPractice Quiz

1.Given the reaction:

Al4C3 (s) + 12 H2O (l) = 4 Al(OH)3 (aq) + 3 CH4 (g)

Answer the following:

a) How many mols of water are needed exactly react with 8.75 mols ofAl4C3 (s) ?

b) How many mols of CH4 (g) will be produce by the reaction of 3.50 molsof Al4C3 (s) ?

c) How many grams of CH4 (g) will be produced by the reaction of 6.00mols of H20 (l) ?

d) How many grams of Al4C3 (s) need to react to produce 75.0 grams ofAl(OH)3 (aq) ?

2. (See Problem 17, Text,page 246.) Methanol, CH3OH (l), can be synthesizedfrom the direct reaction of carbon monoxide gas and hydrogen.

CO (g) + 2 H2 (g) = CH3OH (l)

a) Starting with a mixture containing 12.0 grams of H2 (g) and 74.5 gramsof CO, find the limiting reactant.

b) What is the theoretical yield of methanol from this reaction mixturein grams ?

c) How many grams of the excess reactant remain unreacted after the reactionis complete ?

d) If the actual yield of methanol from the mixture was found to be 52.35grams, what as the percentage yield of this reaction?

Answers:

1 a) 8.75 mols Al4C3 x [12 mols H2O / 1 mol Al4C3 ] = 105 mols of water

b) 3.50 mols Al4C3 x [3 mols CH4 / 1 mol Al4C3 ] = 10.50 mols of methane

c) 6.00 mols H2O x [3 mols CH4 / 12 mols H2O ] = 1.50 mols CH4 x 16.0gm/mol = 24.0 gms

d) 75.0 gms Al(OH)3 x [1 mol / 78 gms ] = 0.962 mols Al(OH)3

0.962 mols Al(OH)3 x [ 1 mol Al4C3 / 4 mol Al(OH)3 ] = 0.240 mols ofAl4C3 needed

0.240 mols x 144 gm/mol = 34.56 grams of Al4C3

2. Actual mol ratio based on masses of reactants given is 6.0 mols H2to 2.66 mols CO or after dividing H2 /CO

2.25 mols of H2 / 1.00 mol CO compare this to 2.00 mols H2 / 1.00 mol CO

Since the numerator "H2" in the actual case is greater thanthe stoichiometric factor of 2.00 mols. hydrogen gas is in excess and thelimiting reactant is 74.5 grams or 2.66 mols of CO (g).

b) 2.66 mols CO x [1 mol CH4 / 1 mol CO ] = 2.66 mols CH3OH produced

2.66 mols x 32.0 gm/mol = 85.12 grams as the theoretical yield of methanol.

c) You need 2.66 mol CO x [2 mol H2 / 1 mol CO ] = 5.32 mol H2 (g) toreact with the CO gas present. Since you atarted with 6.00 mols of H2, 6.00mols - 5.32 mols = 0.68 mols H2 (g) left unreacted. 0.68 mols H2 x 2.00gm/mol = 1.36 grams of H2 remaining.

d) Percent yield = 52.35 grams / 85.12 grams x 100% = 61.50 % yield.