5 Stoichiometry Problems Solved: Answer Key Included

In the fascinating world of chemistry, one subject that often stands out is stoichiometry. This field is essentially the study of quantitative relationships in chemical reactions, providing a bridge between the macroscopic (the lab experiments you might perform) and the microscopic (the molecular level). Understanding stoichiometry is crucial for anyone delving into chemical sciences or related fields, as it provides insight into how chemicals react, how much product is expected, or how much of a reactant is needed. In this detailed exploration, we will solve five fundamental stoichiometry problems to enhance your understanding. Each problem will be dissected step-by-step, with answers provided at the end for your reference.

Problem 1: Calculating Moles from Mass

Given the balanced equation:

2Na(s) + Cl₂(g) → 2NaCl(s)

Calculate the number of moles of sodium chloride (NaCl) that can be formed when 2.00 grams of sodium (Na) reacts with chlorine gas (Cl₂).

• Step 1: Find the molar mass of sodium:

Na has an atomic mass of approximately 22.99 g/mol.

• Step 2: Convert grams of Na to moles:

\[ \text{moles of Na} = \frac{2.00 \text{ g}}{22.99 \text{ g/mol}} = 0.0870 \text{ mol} \]

• Step 3: Use the stoichiometry from the balanced equation:

The equation states 2 moles of Na form 2 moles of NaCl, so:

\[ \text{moles of NaCl} = 0.0870 \text{ mol} \]

Problem 2: Determining the Limiting Reactant

Using the equation:

N₂(g) + 3H₂(g) → 2NH₃(g)

If 1.00 moles of nitrogen (N₂) react with 1.50 moles of hydrogen (H₂), identify the limiting reactant and the amount of ammonia (NH₃) that will be formed.

• Step 1: Determine moles of NH₃ from N₂:

1 mole of N₂ produces 2 moles of NH₃. Thus:

\[ \text{moles of NH}_3 = 1.00 \text{ mol} \times 2 = 2.00 \text{ mol} \]

• Step 2: Determine moles of NH₃ from H₂:

3 moles of H₂ produces 2 moles of NH₃. So, for 1.50 moles:

\[ \text{moles of NH}_3 = \frac{1.50 \text{ mol}}{3} \times 2 = 1.00 \text{ mol} \]

• Step 3: Identify the limiting reactant:

H₂ will be the limiting reactant because it produces less NH₃.

\[ \text{Amount of NH}_3 \text{ formed} = 1.00 \text{ mol} \]

Problem 3: Percent Yield Calculation

Consider the reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

If 5.00 grams of iron (Fe) reacts with excess oxygen (O₂) and yields 3.80 grams of iron(III) oxide (Fe₂O₃), calculate the percent yield of the reaction.

• Step 1: Find the moles of Fe:

The molar mass of Fe is 55.84 g/mol:

\[ \text{moles of Fe} = \frac{5.00 \text{ g}}{55.84 \text{ g/mol}} = 0.0896 \text{ mol} \]

• Step 2: Calculate the theoretical moles of Fe₂O₃:

4 moles of Fe produce 2 moles of Fe₂O₃:

\[ \text{moles of Fe}_2 \text{O}_3 = 0.0896 \text{ mol} \times \frac{2}{4} = 0.0448 \text{ mol} \]

• Step 3: Convert moles of Fe₂O₃ to grams:

Fe₂O₃ has a molar mass of 159.70 g/mol:

\[ \text{Theoretical grams} = 0.0448 \text{ mol} \times 159.70 \text{ g/mol} = 7.15 \text{ g} \]

• Step 4: Calculate percent yield:

\[ \text{Percent yield} = \frac{3.80 \text{ g}}{7.15 \text{ g}} \times 100\% \approx 53.16\% \]

💡 Note: Percent yield calculations help chemists understand reaction efficiency.

Problem 4: Mass-to-Mass Stoichiometry

Given the equation:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

Determine the mass of calcium chloride (CaCl₂) produced when 10.0 grams of calcium carbonate (CaCO₃) reacts completely.

• Step 1: Molar mass of CaCO₃:

CaCO₃ has a molar mass of 100.09 g/mol:

\[ \text{moles of CaCO}_3 = \frac{10.0 \text{ g}}{100.09 \text{ g/mol}} = 0.0999 \text{ mol} \]

• Step 2: Stoichiometry:

1 mole of CaCO₃ produces 1 mole of CaCl₂.

• Step 3: Convert moles of CaCl₂ to grams:

CaCl₂ has a molar mass of 111.0 g/mol:

\[ \text{Mass of CaCl}_2 = 0.0999 \text{ mol} \times 111.0 \text{ g/mol} = 11.09 \text{ g} \]

⚙️ Note: Mass-to-mass stoichiometry is crucial for predicting product quantities in reactions.

Problem 5: Gas Law Stoichiometry

Consider the reaction:

2NO(g) + O₂(g) → 2NO₂(g)

What volume of NO₂ gas at STP (Standard Temperature and Pressure) would be produced from 5.00 L of NO gas at STP?

• Step 1: Use the stoichiometry of the reaction:

2 moles of NO produce 2 moles of NO₂.

• Step 2: Calculate moles of NO at STP:

1 mole of any gas at STP occupies 22.4 L:

\[ \text{moles of NO} = \frac{5.00 \text{ L}}{22.4 \text{ L/mol}} = 0.223 \text{ mol} \]

• Step 3: Volume of NO₂:

Since the ratio is 1:1:

\[ \text{Volume of NO}_2 = 0.223 \text{ mol} \times 22.4 \text{ L/mol} = 5.00 \text{ L} \]

Throughout this detailed walkthrough, we've tackled five different stoichiometry problems to show how fundamental principles like balancing equations, converting between moles, grams, and volumes, and using stoichiometry ratios can answer quantitative questions in chemistry. Whether you're calculating the moles of a product from a given mass of reactant, determining the limiting reactant, or even computing percent yield, these steps are essential. Understanding these processes not only helps in academic settings but also in real-world applications like industrial chemistry, environmental science, and pharmaceutical manufacturing where precise chemical reactions are paramount.

What is stoichiometry?

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Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It’s used to predict how substances will react and what amounts are needed or formed.

Why do chemists calculate percent yield?

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Percent yield is calculated to assess the efficiency of a chemical reaction. It helps in understanding how much of the potential product was actually produced versus the theoretical maximum, which can point to issues like side reactions or incomplete reactions.

How can one determine the limiting reactant in a reaction?

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The limiting reactant is the one that will run out first, thereby limiting the amount of product formed. It’s determined by calculating how much product each reactant would make if they reacted completely and comparing these amounts to find the smallest.