Limiting reagents worksheet part 2 - Free Printable

Problem Analysis and Solution

The image contains two main chemical reaction problems. Let's solve them step by step.

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#### Problem 1: Vanadium (IV) Phosphide Formation

Given:
- Vanadium (V) reacts with phosphorus (P) to form vanadium (IV) phosphide (\( V_3P_4 \)).
- The balanced chemical equation is:
\[
3V + 4P \rightarrow V_3P_4
\]
- We are given:
- 175 grams of vanadium (V)
- 100 grams of phosphorus (P)

We need to:
1. Determine how many grams of \( V_3P_4 \) can be produced.
2. Identify the limiting reactant and calculate the amount of excess reactant left over.

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##### Step 1: Calculate the moles of vanadium (V) and phosphorus (P)

- Molar mass of vanadium (V): \( 51 \, \text{g/mol} \)
- Molar mass of phosphorus (P): \( 31 \, \text{g/mol} \)

\[
\text{Moles of V} = \frac{\text{Mass of V}}{\text{Molar mass of V}} = \frac{175 \, \text{g}}{51 \, \text{g/mol}} \approx 3.43 \, \text{mol}
\]

\[
\text{Moles of P} = \frac{\text{Mass of P}}{\text{Molar mass of P}} = \frac{100 \, \text{g}}{31 \, \text{g/mol}} \approx 3.23 \, \text{mol}
\]

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##### Step 2: Determine the limiting reactant

From the balanced equation:
\[
3V + 4P \rightarrow V_3P_4
\]
- 3 moles of V react with 4 moles of P.

To find the required moles of P for the given moles of V:
\[
\text{Required moles of P} = \left( \frac{4}{3} \right) \times \text{Moles of V} = \left( \frac{4}{3} \right) \times 3.43 \approx 4.57 \, \text{mol}
\]

Since we only have 3.23 moles of P, phosphorus (P) is the limiting reactant.

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##### Step 3: Calculate the theoretical yield of \( V_3P_4 \)

Using the limiting reactant (P):
- From the balanced equation, 4 moles of P produce 1 mole of \( V_3P_4 \).

\[
\text{Moles of } V_3P_4 = \frac{\text{Moles of P}}{4} = \frac{3.23 \, \text{mol}}{4} \approx 0.8075 \, \text{mol}
\]

- Molar mass of \( V_3P_4 \):
\[
\text{Molar mass of } V_3P_4 = (3 \times 51) + (4 \times 31) = 153 + 124 = 277 \, \text{g/mol}
\]

\[
\text{Mass of } V_3P_4 = \text{Moles of } V_3P_4 \times \text{Molar mass of } V_3P_4 = 0.8075 \, \text{mol} \times 277 \, \text{g/mol} \approx 223.39 \, \text{g}
\]

So, the theoretical yield of \( V_3P_4 \) is approximately 233.39 grams.

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##### Step 4: Calculate the excess reactant (vanadium, V)

From the balanced equation, 3 moles of V are required for every 4 moles of P. Since P is the limiting reactant, we use the moles of P to determine how much V is consumed:

\[
\text{Moles of V consumed} = \left( \frac{3}{4} \right) \times \text{Moles of P} = \left( \frac{3}{4} \right) \times 3.23 \approx 2.42 \, \text{mol}
\]

- Initial moles of V: 3.43 mol
- Moles of V left over:
\[
\text{Moles of V left over} = 3.43 \, \text{mol} - 2.42 \, \text{mol} = 1.01 \, \text{mol}
\]

- Mass of V left over:
\[
\text{Mass of V left over} = \text{Moles of V left over} \times \text{Molar mass of V} = 1.01 \, \text{mol} \times 51 \, \text{g/mol} \approx 51.61 \, \text{g}
\]

So, the excess reactant (vanadium) left over is approximately 51.61 grams.

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#### Final Answers for Problem 1:
1. Grams of \( V_3P_4 \) produced: \(\boxed{233.39 \, \text{g}}\)
2. Excess reactant left over: \(\boxed{51.61 \, \text{g}}\)

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#### Problem 2: Bismuth (III) Chloride and Fluorine Gas Reaction

This problem involves a single replacement reaction between bismuth (III) chloride (\( BiCl_3 \)) and fluorine gas (\( F_2 \)). However, the image does not provide enough information to complete this part. If you need help with this part, please provide additional details or clarify the question.

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#### Summary of Final Answers:
1. Vanadium (IV) Phosphide Formation:
- Grams of \( V_3P_4 \) produced: \(\boxed{233.39 \, \text{g}}\)
- Excess reactant left over: \(\boxed{51.61 \, \text{g}}\)