A Problem Using the Combined Concepts of Stoichiometry (3.3.6) Flashcards
• Several tools can be brought together to solve a stoichiometry problem.
• Several tools can be brought together to solve a stoichiometry problem.
Problem: 7.75 g of aluminum, 8.00 g of sodium
hydroxide, and 100 g of water were reacted.
0.541 g of hydrogen gas was collected. What was
the percent yield?
First, the chemical equation must be balanced. In
the balanced chemical equation, 2 mol Al react with
2 mol NaOH and 6 mol H2O to produce 2 mol
NaAl(OH)4 and 3 mol H2.
Second, the masses of reagents given must be
converted to moles. The molar mass of each
reactant is multiplied by the mass of each reactant
to give moles.
For example, 7.75 g Al is equivalent to 0.287 mol Al.
Next, the limiting reagent must be determined.
To determine the limiting reagent, calculate how
much hydrogen gas could be produced from each
reactant. The reactant that results in the lowest
number of moles of hydrogen gas is the limiting
reagent. In this problem, NaOH is the limiting
reagent.
Next, the theoretical yield must be determined.
Using the reactants given, 0.300 mol H2 can be
produced. This is the theoretical yield of hydrogen
gas.
Finally, the percent yield must be calculated. The
percent yield is the actual yield divided by the
theoretical yield and multiplied by 100%. In this
problem, the actual yield is
0.541 g H2 • 1 mol H2 / 2.0158 g H2 = 0.268 mol H2.
The percent yield is 89.3%. This percent yield
means that 89.3% of the hydrogen gas that could
have been produced was collected.
Problem: 7.75 g of aluminum, 8.00 g of sodium
hydroxide, and 100 g of water were reacted.
0.541 g of hydrogen gas was collected. What was
the percent yield?
First, the chemical equation must be balanced. In
the balanced chemical equation, 2 mol Al react with
2 mol NaOH and 6 mol H2O to produce 2 mol
NaAl(OH)4 and 3 mol H2.
Second, the masses of reagents given must be
converted to moles. The molar mass of each
reactant is multiplied by the mass of each reactant
to give moles.
For example, 7.75 g Al is equivalent to 0.287 mol Al.
Next, the limiting reagent must be determined.
To determine the limiting reagent, calculate how
much hydrogen gas could be produced from each
reactant. The reactant that results in the lowest
number of moles of hydrogen gas is the limiting
reagent. In this problem, NaOH is the limiting
reagent.
Next, the theoretical yield must be determined.
Using the reactants given, 0.300 mol H2 can be
produced. This is the theoretical yield of hydrogen
gas.
Finally, the percent yield must be calculated. The
percent yield is the actual yield divided by the
theoretical yield and multiplied by 100%. In this
problem, the actual yield is
0.541 g H2 • 1 mol H2 / 2.0158 g H2 = 0.268 mol H2.
The percent yield is 89.3%. This percent yield
means that 89.3% of the hydrogen gas that could
have been produced was collected.
