Limiting Reagent Worksheet 1 - Free Printable

Problem Analysis:

We are tasked with solving two limiting reagent problems. For each reaction, we need to:
1. Identify the limiting reagent.
2. Determine the maximum amount of each product that can be formed.
3. Calculate how much of the excess reagent remains after the reaction is complete.

Let's solve each problem step by step.

---

Problem 1:

#### Reaction:
$$
3 \text{NH}_4\text{NO}_3 + \text{Na}_3\text{PO}_4 \rightarrow (\text{NH}_4)_3\text{PO}_4 + 3 \text{NaNO}_3
$$

#### Given:
- 30 grams of ammonium nitrate ($\text{NH}_4\text{NO}_3$)
- 50 grams of sodium phosphate ($\text{Na}_3\text{PO}_4$)

#### Step 1: Write the balanced chemical equation.
The equation is already balanced:
$$
3 \text{NH}_4\text{NO}_3 + \text{Na}_3\text{PO}_4 \rightarrow (\text{NH}_4)_3\text{PO}_4 + 3 \text{NaNO}_3
$$

#### Step 2: Calculate the molar masses of the reactants.
- Molar mass of $\text{NH}_4\text{NO}_3$:
$$
\text{N} = 14, \, \text{H} = 1, \, \text{O} = 16
$$
$$
\text{Molar mass of } \text{NH}_4\text{NO}_3 = (2 \times 14) + (4 \times 1) + (3 \times 16) = 28 + 4 + 48 = 80 \, \text{g/mol}
$$

- Molar mass of $\text{Na}_3\text{PO}_4$:
$$
\text{Na} = 23, \, \text{P} = 31, \, \text{O} = 16
$$
$$
\text{Molar mass of } \text{Na}_3\text{PO}_4 = (3 \times 23) + 31 + (4 \times 16) = 69 + 31 + 64 = 164 \, \text{g/mol}
$$

#### Step 3: Convert the given masses to moles.
- Moles of $\text{NH}_4\text{NO}_3$:
$$
\text{Moles of } \text{NH}_4\text{NO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{30 \, \text{g}}{80 \, \text{g/mol}} = 0.375 \, \text{mol}
$$

- Moles of $\text{Na}_3\text{PO}_4$:
$$
\text{Moles of } \text{Na}_3\text{PO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{50 \, \text{g}}{164 \, \text{g/mol}} \approx 0.305 \, \text{mol}
$$

#### Step 4: Determine the stoichiometric ratio.
From the balanced equation:
$$
3 \text{NH}_4\text{NO}_3 : 1 \text{Na}_3\text{PO}_4
$$
This means 3 moles of $\text{NH}_4\text{NO}_3$ react with 1 mole of $\text{Na}_3\text{PO}_4$.

#### Step 5: Identify the limiting reagent.
- Based on the stoichiometry, 3 moles of $\text{NH}_4\text{NO}_3$ require 1 mole of $\text{Na}_3\text{PO}_4$.
- We have 0.375 moles of $\text{NH}_4\text{NO}_3$ and 0.305 moles of $\text{Na}_3\text{PO}_4$.
- To find out how much $\text{Na}_3\text{PO}_4$ is needed for 0.375 moles of $\text{NH}_4\text{NO}_3$:
$$
\text{Moles of } \text{Na}_3\text{PO}_4 \text{ required} = \frac{0.375}{3} = 0.125 \, \text{mol}
$$
Since we have 0.305 moles of $\text{Na}_3\text{PO}_4$, which is more than 0.125 moles, $\text{NH}_4\text{NO}_3$ is the limiting reagent.

#### Step 6: Calculate the maximum amount of products formed.
- From the balanced equation, 3 moles of $\text{NH}_4\text{NO}_3$ produce 1 mole of $(\text{NH}_4)_3\text{PO}_4$ and 3 moles of $\text{NaNO}_3$.
- Using the limiting reagent ($\text{NH}_4\text{NO}_3$):
- Moles of $(\text{NH}_4)_3\text{PO}_4$ produced:
$$
\text{Moles of } (\text{NH}_4)_3\text{PO}_4 = \frac{0.375}{3} = 0.125 \, \text{mol}
$$
- Moles of $\text{NaNO}_3$ produced:
$$
\text{Moles of } \text{NaNO}_3 = 0.375 \, \text{mol}
$$

- Calculate the masses of the products:
- Molar mass of $(\text{NH}_4)_3\text{PO}_4$:
$$
\text{Molar mass of } (\text{NH}_4)_3\text{PO}_4 = (3 \times 14) + (12 \times 1) + 31 + (4 \times 16) = 42 + 12 + 31 + 64 = 149 \, \text{g/mol}
$$
$$
\text{Mass of } (\text{NH}_4)_3\text{PO}_4 = 0.125 \, \text{mol} \times 149 \, \text{g/mol} = 18.625 \, \text{g}
$$

- Molar mass of $\text{NaNO}_3$:
$$
\text{Molar mass of } \text{NaNO}_3 = 23 + 14 + (3 \times 16) = 23 + 14 + 48 = 85 \, \text{g/mol}
$$
$$
\text{Mass of } \text{NaNO}_3 = 0.375 \, \text{mol} \times 85 \, \text{g/mol} = 31.875 \, \text{g}
$$

#### Step 7: Calculate the excess reagent.
- Moles of $\text{Na}_3\text{PO}_4$ used:
$$
\text{Moles of } \text{Na}_3\text{PO}_4 \text{ used} = 0.125 \, \text{mol}
$$
- Moles of $\text{Na}_3\text{PO}_4$ remaining:
$$
\text{Moles of } \text{Na}_3\text{PO}_4 \text{ remaining} = 0.305 - 0.125 = 0.180 \, \text{mol}
$$
- Mass of $\text{Na}_3\text{PO}_4$ remaining:
$$
\text{Mass of } \text{Na}_3\text{PO}_4 \text{ remaining} = 0.180 \, \text{mol} \times 164 \, \text{g/mol} = 29.52 \, \text{g}
$$

#### Final Answers for Problem 1:
$$
\boxed{
\begin{aligned}
&\text{a) Limiting reagent: } \text{NH}_4\text{NO}_3 \\
&\text{b) Maximum amounts of products: } \\
&\quad (\text{NH}_4)_3\text{PO}_4 = 18.625 \, \text{g}, \, \text{NaNO}_3 = 31.875 \, \text{g} \\
&\text{c) Excess reagent left: } \text{Na}_3\text{PO}_4 = 29.52 \, \text{g}
\end{aligned}
}
$$

---

Problem 2:

#### Reaction:
$$
3 \text{CaCO}_3 + 2 \text{FePO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{Fe}_2(\text{CO}_3)_3
$$

#### Given:
- 100 grams of calcium carbonate ($\text{CaCO}_3$)
- 45 grams of iron(III) phosphate ($\text{FePO}_4$)

#### Step 1: Write the balanced chemical equation.
The equation is already balanced:
$$
3 \text{CaCO}_3 + 2 \text{FePO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{Fe}_2(\text{CO}_3)_3
$$

#### Step 2: Calculate the molar masses of the reactants.
- Molar mass of $\text{CaCO}_3$:
$$
\text{Ca} = 40, \, \text{C} = 12, \, \text{O} = 16
$$
$$
\text{Molar mass of } \text{CaCO}_3 = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100 \, \text{g/mol}
$$

- Molar mass of $\text{FePO}_4$:
$$
\text{Fe} = 56, \, \text{P} = 31, \, \text{O} = 16
$$
$$
\text{Molar mass of } \text{FePO}_4 = 56 + 31 + (4 \times 16) = 56 + 31 + 64 = 151 \, \text{g/mol}
$$

#### Step 3: Convert the given masses to moles.
- Moles of $\text{CaCO}_3$:
$$
\text{Moles of } \text{CaCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{100 \, \text{g}}{100 \, \text{g/mol}} = 1.0 \, \text{mol}
$$

- Moles of $\text{FePO}_4$:
$$
\text{Moles of } \text{FePO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{45 \, \text{g}}{151 \, \text{g/mol}} \approx 0.298 \, \text{mol}
$$

#### Step 4: Determine the stoichiometric ratio.
From the balanced equation:
$$
3 \text{CaCO}_3 : 2 \text{FePO}_4
$$
This means 3 moles of $\text{CaCO}_3$ react with 2 moles of $\text{FePO}_4$.

#### Step 5: Identify the limiting reagent.
- Based on the stoichiometry, 3 moles of $\text{CaCO}_3$ require 2 moles of $\text{FePO}_4$.
- We have 1.0 mole of $\text{CaCO}_3$ and 0.298 moles of $\text{FePO}_4$.
- To find out how much $\text{FePO}_4$ is needed for 1.0 mole of $\text{CaCO}_3$:
$$
\text{Moles of } \text{FePO}_4 \text{ required} = \frac{2}{3} \times 1.0 = 0.667 \, \text{mol}
$$
Since we have only 0.298 moles of $\text{FePO}_4$, which is less than 0.667 moles, $\text{FePO}_4$ is the limiting reagent.

#### Step 6: Calculate the maximum amount of products formed.
- From the balanced equation, 2 moles of $\text{FePO}_4$ produce 1 mole of $\text{Ca}_3(\text{PO}_4)_2$ and 1 mole of $\text{Fe}_2(\text{CO}_3)_3$.
- Using the limiting reagent ($\text{FePO}_4$):
- Moles of $\text{Ca}_3(\text{PO}_4)_2$ produced:
$$
\text{Moles of } \text{Ca}_3(\text{PO}_4)_2 = \frac{0.298}{2} = 0.149 \, \text{mol}
$$
- Moles of $\text{Fe}_2(\text{CO}_3)_3$ produced:
$$
\text{Moles of } \text{Fe}_2(\text{CO}_3)_3 = \frac{0.298}{2} = 0.149 \, \text{mol}
$$

- Calculate the masses of the products:
- Molar mass of $\text{Ca}_3(\text{PO}_4)_2$:
$$
\text{Molar mass of } \text{Ca}_3(\text{PO}_4)_2 = (3 \times 40) + 2 \times (31 + 4 \times 16) = 120 + 2 \times (31 + 64) = 120 + 2 \times 95 = 120 + 190 = 310 \, \text{g/mol}
$$
$$
\text{Mass of } \text{Ca}_3(\text{PO}_4)_2 = 0.149 \, \text{mol} \times 310 \, \text{g/mol} = 46.19 \, \text{g}
$$

- Molar mass of $\text{Fe}_2(\text{CO}_3)_3$:
$$
\text{Molar mass of } \text{Fe}_2(\text{CO}_3)_3 = (2 \times 56) + 3 \times (12 + 3 \times 16) = 112 + 3 \times (12 + 48) = 112 + 3 \times 60 = 112 + 180 = 292 \, \text{g/mol}
$$
$$
\text{Mass of } \text{Fe}_2(\text{CO}_3)_3 = 0.149 \, \text{mol} \times 292 \, \text{g/mol} = 43.51 \, \text{g}
$$

#### Step 7: Calculate the excess reagent.
- Moles of $\text{CaCO}_3$ used:
$$
\text{Moles of } \text{CaCO}_3 \text{ used} = \frac{3}{2} \times 0.298 = 0.447 \, \text{mol}
$$
- Moles of $\text{CaCO}_3$ remaining:
$$
\text{Moles of } \text{CaCO}_3 \text{ remaining} = 1.0 - 0.447 = 0.553 \, \text{mol}
$$
- Mass of $\text{CaCO}_3$ remaining:
$$
\text{Mass of } \text{CaCO}_3 \text{ remaining} = 0.553 \, \text{mol} \times 100 \, \text{g/mol} = 55.3 \, \text{g}
$$

#### Final Answers for Problem 2:
$$
\boxed{
\begin{aligned}
&\text{a) Limiting reagent: } \text{FePO}_4 \\
&\text{b) Maximum amounts of products: } \\
&\quad \text{Ca}_3(\text{PO}_4)_2 = 46.19 \, \text{g}, \, \text{Fe}_2(\text{CO}_3)_3 = 43.51 \, \text{g} \\
&\text{c) Excess reagent left: } \text{CaCO}_3 = 55.3 \, \text{g}
\end{aligned}
}
$$

---

Final Answer:

$$
\boxed{
\begin{aligned}
&\text{Problem 1: } \\
&\quad \text{a) Limiting reagent: } \text{NH}_4\text{NO}_3 \\
&\quad \text{b) Products: } (\text{NH}_4)_3\text{PO}_4 = 18.625 \, \text{g}, \, \text{NaNO}_3 = 31.875 \, \text{g} \\
&\quad \text{c) Excess reagent: } \text{Na}_3\text{PO}_4 = 29.52 \, \text{g} \\
&\text{Problem 2: } \\
&\quad \text{a) Limiting reagent: } \text{FePO}_4 \\
&\quad \text{b) Products: } \text{Ca}_3(\text{PO}_4)_2 = 46.19 \, \text{g}, \, \text{Fe}_2(\text{CO}_3)_3 = 43.51 \, \text{g} \\
&\quad \text{c) Excess reagent: } \text{CaCO}_3 = 55.3 \, \text{g}
\end{aligned}
}
$$