7: Stoichiometry

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Stoichiometric Reactions

Exercise \(\PageIndex{1.1}\)

For the balanced chemical equation

Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s)

how many moles and grams of Ag are produced when 0.661 mol of Al are reacted?

Answer

\(0.661 mol Al*\left ( \frac{3 mol Ag}{1 mol Al} \right )= 1.98 mol Ag\)

\(0.661 mol Al*\left ( \frac{3 mol Ag}{1 mol Al} \right )*\left ( \frac{107.8682 g Ag}{1 mol Ag} \right )=214 grams Ag\)

Exercise \(\PageIndex{1.2}\)

For the balanced chemical reaction

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(ℓ)

how many moles and grams of H₂O are produced when 0.669 mol of NH₃ react?

Answer

\[.669molNH_3\left ( \frac{6molH_2O}{4molNH_3} \right )=1.00molH_2O\]

1.00 mol H₂O, 18.0 grams of H₂O

Exercise \(\PageIndex{1.3}\)

For the balanced chemical reaction

4NaOH(aq) + 2S(s) + 3O₂(g) → 2Na₂SO₄(aq) + 2H₂O(ℓ)

how many moles and grams of Na₂SO₄ are formed when 1.22 grams of O₂ react?

Answer

\(1.22 g O_{2}*\left ( \frac{1 mol O_{2}}{31.988 g O_{2}} \right )*\left ( \frac{2 mol Na_{2}SO_{4}}{3 mol O_{2}} \right )= 0.0254 mol Na_{2}SO_{4}\)

\(1.22 g O_{2}*\left ( \frac{1 mol O_{2}}{31.988 g O_{2}} \right )*\left ( \frac{2 mol Na_{2}SO_{4}}{3 mol O_{2}} \right )*\left ( \frac{142.04314 g Na_{2}SO_{4}}{1 mol Na_{2}SO_{4}} \right )=3.61 g Na_{2}SO_{4}\)

Exercise \(\PageIndex{1.4}\)

For the balanced chemical reaction

4KO₂(s) + 2CO₂(g) → 2K₂CO₃(s) + 3O₂(g)

determine the number of moles and grams of both products formed when 6.88 grams of KO₂ react.

Answer

Moles: 0.0484 mol K₂CO₃, 0.0726 mol O₂

Grams: 6.69 grams K₂CO₃, 2.32 grams O₂

Theoretical Yield, Limiting and Excess Reagents

Exercise \(\PageIndex{2.1}\)

The box below shows a group of nitrogen and hydrogen molecules that will react to produce ammonia, NH₃. What is the limiting reagent?

Answer: Nitrogen is the limiting reagent.

Exercise \(\PageIndex{2.2}\)

Given the statement “20.0 g of methane is burned in excess oxygen,” is it obvious which reactant is the limiting reagent?

Answer: Yes; methane is the limiting reagent.

Exercise \(\PageIndex{2.3}\)

Acetylene (C₂H₂) is formed by reacting 7.08 g of C and 4.92 g of H₂.

2C(s) + H₂(g) → C₂H₂(g)

What is the limiting reagent? How much of the other reactant is in excess?

Answer: C is the limiting reagent; 4.33 g of H₂ are left over.

Exercise \(\PageIndex{2.4}\)

To form the precipitate PbCl₂, 2.88 g of NaCl and 7.21 g of Pb(NO₃)₂ are mixed in solution. How much precipitate is formed? How much of which reactant is in excess?

Answer: 6.06 g of PbCl₂ are formed; 0.33 g of NaCl is left over.

Percent Yield

Exercise \(\PageIndex{3.1}\)

What is the difference between the theoretical yield and the actual yield?
What is the difference between the actual yield and the percent yield?

Answer

Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.
Actual yield is what you actually get from a chemical reaction; percent yield is the percent ratio of actual yield to the theoretical yield, meaning it is a ratio that determines the accuracy of the actual yield from an experiment.

Exercise \(\PageIndex{3.2}\)

A worker isolates 2.675 g of SiF₄ after reacting 2.339 g of SiO₂ with HF. What are the theoretical yield and the actual yield? What is the percent yield?

SiO₂(s) + 4HF(g) → SiF₄(g) + 2H₂O(ℓ)

Answer

theoretical yield = 4.052 g; actual yield = 2.675 g

percent yield = 66.02%

Exercise \(\PageIndex{3.3}\)

A chemist decomposes 1.006 g of NaHCO₃ and obtains 0.0334 g of Na₂CO₃. What are the theoretical yield and the actual yield? What is the percent yield?

2NaHCO₃(s) → Na₂CO₃(s) + H₂O(ℓ) + CO₂(g)

Answer

theoretical yield = 0.635 g; actual yield = 0.0334 g

percent yield = 5.26%

Mole to Mole Conversion

Exercise \(\PageIndex{3.3}\)

Analine (C₆H₅NH₂) can be fromend from nitro benzene (C₆H₅NO₂) by the following equation:

4C₆H₅NO₂ + 9Fe + 4H₂O ---> 4C₆H₅NH₂ + 3 Fe₃O₄

1.a) Write the conversion factor converting moles iron to moles analine.

1.b) How many moles of analine would be formed if 3.78 moles of iron was consumed?

1.c )Write the conversion factor converting moles nitro benzene to moles Fe₃O₄.

1.d) Write the conversion factor describing the ration fo the consumption of iron to the consumption of nitrobenzene.

1.e) How many moles of nitrobenzene would be needed to produce 4.678 Mmoles of Fe₃O₄?

1.f) How many moles of iron would be needed to consume 3.56x10²² molecules of nitrobenzene?

1.g) How many atoms of iron are consumed if 3.59nmoles of water are consumed?

Answer 1.a

\((\frac{4\;mol\;C_6H_5NH_2}{9\;mol\;Fe})\)

Answer 1.b

\(3.78mol\;Fe\;(\frac{4\;mol\;C_6H_5NH_2}{9\;mol\;Fe})\;=\;1.68\;mol\;C_6H_5NH_2\)

Answer 1.c

\((\frac{3\;mol\;Fe_3O_4}{4\;mol\;C_6H_5NO_2})\)

Answer 1.d

\((\frac{9\;mol\;Fe}{4\;mol\;C_6H_5NO_2})\)

Answer 1.e

\(4.678Mmole\;Fe_3O_4(\frac{10^6\;mole\;Fe_3O_4}{Mmol\;Fe_3O_4})(\frac{4\;mol\;C_6H_5NO_2}{3\;mol\;Fe_3O_4})\)

\(=\;6.237\;x\;10^6\;mol\; C_6H_5NO_2\; or \;6.237\;Mmol\;C_6H_5NO_2\)

Answer 1.f

\(3.56\;\times\;10^{22}\;molecule\;C_6H_5NO_2\;(\frac{mol\;C_6H_5NO_2}{6.022\;\times\;10^{23}\;molecule})(\frac{9\;mol\;Fe}{4\;mol\; C_6H_5NO_2 })\)

Answer 1.g

\(3.59\;nmol\;H_2O\;(\frac{10^{-9}mol\;H_2O}{nmol})(\frac{9\;mol\;Fe}{4\;mol\;H_2O})(\frac{6.022\;\times\;10^{23}\;atom\;Fe}{Fe})\;=\;4.86\; \times \;atom\;Fe\)

Exercise \(\PageIndex{3.3}\)

Analine (C₆H₅NH₂) can be fromend from nitro benzene (C₆H₅NO₂) by the following equation:

4C₆H₅NO₂ + 9Fe + 4H₂O ---> 4C₆H₅NH₂ + 3 Fe₃O₄

2.a. How many grams of analine would be formed if 3.78moles of iron was consumed?

2.b. How many grams of nitrobenzene would be needed to produce 4.678 kg of Fe₃O₄?

2.c. How many grams of iron would be needed to consume 3.56x10²² molecules of nitrobenzene?

Answer 1.a

\(3.78\;mol\;Fe\;(\frac{4\;mol\;C_6H_5NH_2}{9\;mol\;Fe})(\frac{93.121g\;C_6H_5NH_2}{mol})\;=\;156g\;C_6H_5NH_2\)

Answer 1.b

\(4.678kg\;Fe_3O_4\;(\frac{10^3\;g\;Fe_3O_4}{kg})(\frac{mol\;Fe_3O_4}{2.31.535\;g\;Fe_3O_4})(\frac{4\;mol\;C_6H_5NH_2}{3\;mol\;Fe_3O_4})(\frac{123.105g\; C_6H_5NH_2}{mol})\)

\(=\;3.316x10^3\;g\; C_6H_5NO_2\;or\;3.316kg\;C_6H_5N)_2 \)

Answer 1.c

\((\frac{3\;mol\;Fe_3O_4}{4\;mol\;C_6H_5NO_2})\)

Exercise \(\PageIndex{3.3}\)

Analine (C₆H₅NH₂) can be fromend from nitro benzene (C₆H₅NO₂) by the following equation:

4C₆H₅NO₂ + 9Fe + 4H₂O ---> 4C₆H₅NH₂ + 3 Fe₃O₄

2.a. How many grams of analine would be formed if 3.78moles of iron was consumed?

2.b. How many grams of nitrobenzene would be needed to produce 4.678 kg of Fe₃O₄?

2.c. How many grams of iron would be needed to consume 3.56x10²² molecules of nitrobenzene?

Answer 1.a

\(3.78\;mol\;Fe\;(\frac{4\;mol\;C_6H_5NH_2}{9\;mol\;Fe})(\frac{93.121g\;C_6H_5NH_2}{mol})\;=\;156g\;C_6H_5NH_2\)

Answer 1.b

\(4.678kg\;Fe_3O_4\;(\frac{10^3\;g\;Fe_3O_4}{kg})(\frac{mol\;Fe_3O_4}{2.31.535\;g\;Fe_3O_4})(\frac{4\;mol\;C_6H_5NH_2}{3\;mol\;Fe_3O_4})(\frac{123.105g\; C_6H_5NH_2}{mol})\)

\(=\;3.316x10^3\;g\; C_6H_5NO_2\;or\;3.316kg\;C_6H_5N)_2 \)

Answer 1.c

\((\frac{3\;mol\;Fe_3O_4}{4\;mol\;C_6H_5NO_2})\)