Limiting Reagent Worksheet #2 Notes

Limiting Reagent Problems

Problem 1

The balanced chemical reaction is:

$I<em>2O</em>5(g) + 5CO(g) \longrightarrow 5CO<em>2(g) + I</em>2(g)$

a) Given:

• 80.0 grams of iodine(V) oxide ($I<em>2O</em>5$)

• 28.0 grams of carbon monoxide (CO)

  Task: Determine the mass of iodine ($I_2$) produced.

  Solution Approach:

• Convert the mass of each reactant to moles.

• Determine the limiting reactant by comparing the mole ratios to the balanced equation.

• Calculate the theoretical yield of iodine ($I_2$) based on the limiting reactant.

  Calculations:

  Molar mass of $I<em>2O</em>5$ = 2(126.90) + 5(16.00) = 253.8 + 80.0 = 333.8 g/mol
  Moles of $I<em>2O</em>5$ = $\frac{80.0 \text{ g}}{333.8 \text{ g/mol}} = 0.2397 \text{ mol}$

  Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol
  Moles of CO = $\frac{28.0 \text{ g}}{28.01 \text{ g/mol}} = 0.9996 \text{ mol}$

  From the balanced equation, 1 mole of $I<em>2O</em>5$ reacts with 5 moles of CO.

  Check the mole ratio:
  $\frac{\text{Moles of CO}}{\text{Moles of } I<em>2O</em>5} = \frac{0.9996}{0.2397} = 4.17$ which is less than 5. Therefore, CO is the limiting reactant.

  Based on CO, the moles of $I<em>2$ produced: Since 5 moles of CO produces 1 mole of $I</em>2$,
  Moles of $I<em>2$ = $\frac{0.9996 \text{ mol CO}}{5} = 0.1999 \text{ mol } I</em>2$

  Mass of $I<em>2$ produced = Moles of $I</em>2$ * Molar mass of $I<em>2$ Molar mass of $I</em>2$ = 2 * 126.90 = 253.8 g/mol
  Mass of $I_2$ = 0.1999 mol * 253.8 g/mol = 50.73 grams

  Answer: 50.73 grams of $I_2$ could be produced.

b) Given:

• Actual moles of $I_2$ produced = 0.160 moles

  i) Task: What mass of iodine was produced?

  Mass of $I<em>2$ produced = Moles of $I</em>2$ * Molar mass of $I<em>2$ Mass of $I</em>2$ = 0.160 mol * 253.8 g/mol = 40.61 grams

  Answer: 40.61 grams of iodine was produced.

  ii) Task: What percentage yield of iodine was produced?

  Percentage Yield = $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} * 100$
  Percentage Yield = $\frac{0.160 \text{ mol}}{0.1999 \text{ mol}} * 100 = 80.04 \%$

  Answer: 80.04% yield of iodine was produced.

Problem 2

The balanced chemical reaction is:

$Zn + S \longrightarrow ZnS$

a) Given:

• 25.0 g of zinc (Zn)

• 30.0 g of sulfur (S)

  Task: Which chemical is the limiting reactant?

  Calculations:

  Molar mass of Zn = 65.38 g/mol
  Moles of Zn = $\frac{25.0 \text{ g}}{65.38 \text{ g/mol}} = 0.3824 \text{ mol}$

  Molar mass of S = 32.07 g/mol
  Moles of S = $\frac{30.0 \text{ g}}{32.07 \text{ g/mol}} = 0.9355 \text{ mol}$

  From the balanced equation, 1 mole of Zn reacts with 1 mole of S.

  Check the mole ratio:
  Since the mole ratio is 1:1, comparing the moles directly shows that Zn has fewer moles than S. Therefore, Zn is the limiting reactant.

  Answer: Zinc (Zn) is the limiting reactant.

b) Task: How many grams of ZnS will be formed?

Based on the limiting reactant (Zn), the moles of ZnS produced:
Since 1 mole of Zn produces 1 mole of ZnS,
Moles of ZnS = 0.3824 mol

Mass of ZnS produced = Moles of ZnS * Molar mass of ZnS
Molar mass of ZnS = 65.38 + 32.07 = 97.45 g/mol
Mass of ZnS = 0.3824 mol * 97.45 g/mol = 37.26 grams

Answer: 37.26 grams of ZnS will be formed.

c) Task: How many grams of the excess reactant will remain after the reaction is over?

Moles of S reacted = Moles of Zn reacted = 0.3824 mol
Moles of S remaining = Initial moles of S - Moles of S reacted
Moles of S remaining = 0.9355 mol - 0.3824 mol = 0.5531 mol

Mass of S remaining = Moles of S remaining * Molar mass of S
Mass of S remaining = 0.5531 mol * 32.07 g/mol = 17.74 grams

Answer: 17.74 grams of sulfur will remain after the reaction is over.

Problem 3

Given:

• 3.00 grams of Mg

• 2.20 grams of $O_2$

  The balanced chemical reaction is:

  $2Mg + O_2 \longrightarrow 2MgO$

  Task:

• Which element is in excess?

• What mass is in excess?

• What mass of MgO is formed?

  Calculations:

  Molar mass of Mg = 24.31 g/mol
  Moles of Mg = $\frac{3.00 \text{ g}}{24.31 \text{ g/mol}} = 0.1234 \text{ mol}$

  Molar mass of $O<em>2$ = 2 * 16.00 = 32.00 g/mol Moles of $O</em>2$ = $\frac{2.20 \text{ g}}{32.00 \text{ g/mol}} = 0.06875 \text{ mol}$

  From the balanced equation, 2 moles of Mg react with 1 mole of $O_2$.

  Check the mole ratio:
  $\frac{\text{Moles of Mg}}{\text{Moles of } O<em>2} = \frac{0.1234}{0.06875} = 1.795$ Since the ratio is less than 2, Mg is the limiting reactant, and $O</em>2$ is in excess.

  Answer 1: Oxygen is in excess.

  Moles of $O<em>2$ reacted = 0.5 * Moles of Mg = 0.5 * 0.1234 mol = 0.0617 mol Moles of $O</em>2$ remaining = Initial moles of $O<em>2$ - Moles of $O</em>2$ reacted
  Moles of $O_2$ remaining = 0.06875 mol - 0.0617 mol = 0.00705 mol

  Mass of $O<em>2$ remaining = Moles of $O</em>2$ remaining * Molar mass of $O<em>2$ Mass of $O</em>2$ remaining = 0.00705 mol * 32.00 g/mol = 0.2256 grams

  Answer 2: 0.2256 grams of oxygen is in excess.

  Based on the limiting reactant (Mg), the moles of MgO produced:
  Since 2 moles of Mg produces 2 moles of MgO, 1 mole of Mg produces 1 mole of MgO.
  Moles of MgO = Moles of Mg = 0.1234 mol

  Mass of MgO produced = Moles of MgO * Molar mass of MgO
  Molar mass of MgO = 24.31 + 16.00 = 40.31 g/mol
  Mass of MgO = 0.1234 mol * 40.31 g/mol = 4.97 grams

  Answer 3: 4.97 grams of MgO is formed.