Stoichiometry & Percent Yield Practice2 Al + 6 HCl 2 AlCl3 + 3 H2

95 g of hydrochloric acid are reacted with excess aluminum producing 2.2 g of hydrogen gas. What is the percent yield?

Solution:

23 mol H6 mol HCl

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

1 mol HCl95 g HCl36.46 g HCl⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2

2

2.016 g H1 mol H

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2= 2.6 g H

Calculate theoretical yield:

Next, calculate % yield:

Stoichiometry & Percent Yield Practice2 HgO 2 Hg + O2

42 g of mercury(II) oxide are decomposed. If the % yield is 99.0%, what mass of oxygen gas will actually be produced?

Solution:

21 mol O2 mol HgO

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

1 mol HgO42 g HgO216.59 g HgO⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2

2

32.00 g O1 mol O

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Begin like you are calculating the theoretical yield:

Then, multiply by the % yield:

99100⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠ 2= 3.1 g O

Stoichiometry & Percent Yield Practice2 Na2O2 + 2 H2O 4 NaOH + O2

What mass of sodium peroxide is needed to produce 19 g of oxygen gas if the % yield is 96.0%?

Solution:

2

2

1 mol O32.00 g O

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

210019 g O96

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2 2

2

2 mol Na O1 mol O

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Begin by changing the actual yield into a theoretical yield by dividing by the % yield:

Then, do a typical stoichiometric calculation:

2 2

2 2

77.98 g Na O1 mol Na O

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2 2= 96 g Na O

Stoichiometry & Percent Yield Practice4 Al + 3 O2 2 Al2O3

16 g of aluminum are reacted with excess oxygen gas producing 29 g of aluminum oxide. What is the percent yield?

Solution:

2 32 mol Al O4 mol Al

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

1 mol Al16 g Al26.98 g Al⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2 3

2 3

101.96 g Al O1 mol Al O

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2 3= 30. g Al O

Calculate theoretical yield:

Next, calculate % yield:

2 3

2 3

29 g Al Oactual%Yield = ?100 = ?100 = 96.7 % Yieldtheoretical 30. g Al O

Stoichiometry & Percent Yield PracticeFeCl3 + 3 NH4OH 3 NH4Cl + Fe(OH)3

13 g of iron(III) chloride are reacted with excess ammonium hydroxide. If the % yield is 95.5%, what mass of ammonium chloride will

actually be produced?

Solution:

4

3

3 mol NH Cl1 mol FeCl

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

33

3

1 mol FeCl13 g FeCl

162.2 g FeCl

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

4

4

53.49 g NH Cl1 mol NH Cl

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Begin like you are calculating the theoretical yield:

Then, multiply by the % yield:

95.5100

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

4NH Cl= 12 g

Stoichiometry & Percent Yield Practice2 Na + Cl2 2 NaCl

What mass of chlorine is needed to produce 22 g of sodium chloride if the % yield is 97.0%?

Solution:

1 mol NaCl58.44 g NaCl⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

210022 g O97

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

21 mol Cl2 mol NaCl

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Begin by changing the actual yield into a theoretical yield by dividing by the % yield:

Then, do a typical stoichiometric calculation:

2

2

70.91 g Cl1 mol Cl

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

2= 14 g Cl