Stoichiometry Practice: Step-by-Step Guidance for General Chemistry

Q1. How many moles of Cl2 are required to produce 11.6 moles of PCl3 given the unbalanced reaction: P + Cl2 → PCl3?

Background

Topic: Stoichiometry and Mole Ratios

This question tests your ability to use stoichiometric relationships from a chemical equation to determine the amount of one substance needed to produce a given amount of another.

Key Terms and Formulas

• Stoichiometry: The calculation of reactants and products in chemical reactions.

• Mole Ratio: The ratio between the amounts in moles of any two compounds involved in a chemical reaction.

Step-by-Step Guidance

1. First, balance the chemical equation: $\mathrm{P} + \mathrm{Cl}_2 \rightarrow \mathrm{PCl}_3$

2. Determine the mole ratio between $\mathrm{Cl}_2$ and $\mathrm{PCl}_3$ from the balanced equation.

3. Set up a proportion using the mole ratio to relate moles of $\mathrm{Cl}_2$ to the given moles of $\mathrm{PCl}_3$ (11.6 mol).

4. Write the conversion as: $\text{moles of } \mathrm{Cl}_2 = 11.6 \text{ mol } \mathrm{PCl}_3 \times \left( \frac{\text{mol } \mathrm{Cl}_2}{\text{mol } \mathrm{PCl}_3} \right)$

Try solving on your own before revealing the answer!

Final Answer: 11.6 mol Cl2

Since the balanced equation shows a 1:1 ratio between Cl2 and PCl3, the number of moles required is the same as the moles of PCl3 produced.

Q2. How many grams of each CO2 and NH3 are produced from 0.83 mole of urea? (NH2)2CO + H2O → 2 NH3 + CO2

Background

Topic: Stoichiometry, Mole-to-Mass Conversions

This question tests your ability to use mole ratios and molar masses to convert from moles of a reactant to grams of products.

Key Terms and Formulas

• Mole Ratio: From the balanced equation, 1 mol urea produces 2 mol NH3 and 1 mol CO2.

• Molar Mass: The mass of one mole of a substance (g/mol).

• Mass Calculation: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Step-by-Step Guidance

1. Identify the mole ratios from the balanced equation: 1 mol urea : 2 mol NH3 : 1 mol CO2.

2. Calculate the moles of NH3 and CO2 produced from 0.83 mol urea using the mole ratios.

3. Find the molar masses: $\mathrm{NH}_3$ (17.03 g/mol), $\mathrm{CO}_2$ (44.01 g/mol).

4. Set up the mass calculations: $\text{mass of NH}_3 = \text{moles of NH}_3 \times 17.03$; $\text{mass of CO}_2 = \text{moles of CO}_2 \times 44.01$.

Try solving on your own before revealing the answer!

Final Answer: 1.4 g CO2 and 28.3 g NH3

0.83 mol urea produces 0.83 mol CO2 and 1.66 mol NH3. Multiply by their respective molar masses to get the mass of each product.

Q3. How many grams of N2 are necessary to produce 7.5 g of NH3? N2 + 3 H2 → 2 NH3

Background

Topic: Stoichiometry, Mass-to-Mass Conversions

This question tests your ability to convert between mass of product and mass of reactant using mole ratios and molar masses.

Key Terms and Formulas

• Molar Mass: $\mathrm{NH}_3$ (17.03 g/mol), $\mathrm{N}_2$ (28.02 g/mol)

• Mole Ratio: From the balanced equation, 1 mol N2 produces 2 mol NH3.

• Mass Calculation: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Step-by-Step Guidance

1. Convert 7.5 g NH3 to moles using its molar mass: $\text{moles NH}_3 = \frac{7.5}{17.03}$

2. Use the mole ratio (1 mol N2 : 2 mol NH3) to find moles of N2 needed.

3. Convert moles of N2 to grams using its molar mass: $\text{mass N}_2 = \text{moles N}_2 \times 28.02$

Try solving on your own before revealing the answer!

Final Answer: 6.2 g N2

First, convert grams of NH3 to moles, use the 1:2 ratio to get moles of N2, then multiply by the molar mass of N2.

Q4. What mass (in grams) of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? 2 LiOH (s) + CO2 (g) → Li2CO3 (s) + H2O (l)

Background

Topic: Stoichiometry, Limiting Reactant, Mass-to-Mass Conversion

This question tests your ability to use stoichiometry to determine how much CO2 can react with a given mass of LiOH.

Key Terms and Formulas

• Molar Mass: LiOH (23.95 g/mol), CO2 (44.01 g/mol)

• Mole Ratio: 2 mol LiOH : 1 mol CO2

• Mass Calculation: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

Step-by-Step Guidance

1. Convert 1.00 kg LiOH to grams (1.00 kg = 1000 g).

2. Calculate moles of LiOH: $\text{moles LiOH} = \frac{1000}{23.95}$

3. Use the mole ratio (2:1) to find moles of CO2 that can be absorbed.

4. Convert moles of CO2 to grams: $\text{mass CO}_2 = \text{moles CO}_2 \times 44.01$

Try solving on your own before revealing the answer!

Final Answer: 918 g CO2

After converting LiOH to moles, use the 2:1 ratio to get moles of CO2, then multiply by the molar mass of CO2.

Q5. If 8.60 × 103 g of H2 is reacted and 3.57 × 104 g of CH3OH is actually produced, what is the percent yield of methanol? 2 H2 (g) + CO (g) → CH3OH (l)

Background

Topic: Percent Yield, Stoichiometry

This question tests your ability to calculate theoretical yield and percent yield from given masses of reactant and product.

Key Terms and Formulas

• Theoretical Yield: The maximum amount of product that can be formed from the given amount of reactant.

• Percent Yield: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Molar Masses: H2 (2.02 g/mol), CH3OH (32.05 g/mol)

Step-by-Step Guidance

1. Convert 8.60 × 103 g H2 to moles: $\text{moles H}_2 = \frac{8.60 \times 10^3}{2.02}$

2. Use the mole ratio (2 mol H2 : 1 mol CH3OH) to find moles of CH3OH that can be produced.

3. Convert moles of CH3OH to grams: $\text{theoretical yield} = \text{moles CH}_3\text{OH} \times 32.05$

4. Calculate percent yield: $\text{Percent Yield} = \frac{3.57 \times 10^4}{\text{theoretical yield}} \times 100\%$

Try solving on your own before revealing the answer!

Final Answer: 84.2%

First, find the theoretical yield from the limiting reactant (H2), then use the actual yield to calculate percent yield.

Extra: If 18.1 g of NH3 is reacted with 90.4 g of CuO, how many grams of N2 will be formed? 2 NH3 + 3 CuO → N2 + 3 Cu + 3 H2O

Background

Topic: Limiting Reactant, Stoichiometry, Mass-to-Mass Conversion

This question tests your ability to identify the limiting reactant and use it to determine the amount of product formed.

Key Terms and Formulas

• Limiting Reactant: The reactant that is completely consumed first and limits the amount of product formed.

• Molar Masses: NH3 (17.03 g/mol), CuO (79.55 g/mol), N2 (28.02 g/mol)

• Mole Ratios: 2 mol NH3 : 3 mol CuO : 1 mol N2

Step-by-Step Guidance

1. Convert 18.1 g NH3 and 90.4 g CuO to moles using their molar masses.

2. Determine the limiting reactant by comparing the mole ratios needed for the reaction.

3. Use the limiting reactant to calculate the moles of N2 produced (using the 2:1 or 3:1 ratio as appropriate).

4. Convert moles of N2 to grams: $\text{mass N}_2 = \text{moles N}_2 \times 28.02$

Try solving on your own before revealing the answer!

Final Answer: 8.94 g N2

After finding the limiting reactant, use its moles to determine the amount of N2 formed and convert to grams.