5 Essential Mixed Stoichiometry Problems Solved

In the realm of chemistry, stoichiometry stands as a pivotal concept that connects the mass of reactants to the mass of products in chemical reactions. It's essentially about calculating the quantitative relationships between substances in a given chemical reaction. Today, we delve into the heart of mixed stoichiometry, which involves solving problems where various forms of matter (solids, liquids, gases) interact, often under diverse conditions like temperature and pressure variations. Understanding mixed stoichiometry not only helps in grasping chemical reactions but also equips students with problem-solving skills that are invaluable in various scientific careers.

Problem 1: Gas to Gas Stoichiometry

Let’s start with a problem involving the combustion of ethane gas, a common example where we need to find out how much carbon dioxide gas is produced when a certain amount of ethane is burnt under standard conditions.

Given: - 5.00 grams of ethane (C₂H₆) are combusted in excess oxygen at STP.

Step-by-Step Solution:

• First, we calculate the moles of ethane. The molar mass of C₂H₆ is approximately 30.07 g/mol.
• Moles of C₂H₆ = 5.00 g / 30.07 g/mol ≈ 0.166 mol
• Using the balanced equation: C₂H₆ + ⁷⁄₂ O₂ → 2 CO₂ + 3 H₂O, we see that 1 mole of C₂H₆ produces 2 moles of CO₂.
• Thus, 0.166 mol of C₂H₆ will produce 2 * 0.166 = 0.332 mol of CO₂.
• Convert this back to grams using the molar mass of CO₂, which is 44.01 g/mol:
  • Mass of CO₂ = 0.332 mol * 44.01 g/mol ≈ 14.61 g

💡 Note: Remember to check for significant figures and round your answer appropriately.

Problem 2: Liquid to Solid Stoichiometry

Next, consider the synthesis of silver chloride from silver nitrate solution and sodium chloride solution, focusing on the solid precipitate formed.

Given: - 1.0 L of a 0.2 M silver nitrate (AgNO₃) solution reacts with sodium chloride.

Step-by-Step Solution:

• Calculate the moles of AgNO₃ in the solution:
  • Moles = 1.0 L * 0.2 mol/L = 0.2 mol
• Using the balanced reaction: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq), we see that 1 mol of AgNO₃ produces 1 mol of AgCl.
• Hence, 0.2 mol of AgNO₃ will produce 0.2 mol of AgCl.
• Convert this back to grams using the molar mass of AgCl, which is 143.32 g/mol:
  • Mass of AgCl = 0.2 mol * 143.32 g/mol = 28.66 g

🌟 Note: In this case, the reaction is stoichiometric, so the limiting reactant will be the one in lesser quantity.

Problem 3: Solids to Gas Stoichiometry

Now, let’s examine a reaction involving a solid carbonate and an acid producing a gas:

Given: - 2.50 grams of sodium bicarbonate (NaHCO₃) reacts with excess hydrochloric acid (HCl).

Step-by-Step Solution:

• Calculate the moles of NaHCO₃. The molar mass of NaHCO₃ is about 84.007 g/mol.
• Moles of NaHCO₃ = 2.50 g / 84.007 g/mol ≈ 0.0297 mol
• Using the balanced equation: NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g), 1 mole of NaHCO₃ yields 1 mole of CO₂.
• Therefore, 0.0297 mol of NaHCO₃ will produce 0.0297 mol of CO₂.
• At STP, 1 mol of gas occupies 22.4 L, so:
  • Volume of CO₂ = 0.0297 mol * 22.4 L/mol ≈ 0.665 L

🗒 Note: Volume calculations can be affected by changes in pressure and temperature from STP conditions.

Problem 4: Multiple Gas to Liquid Conversion

Consider a reaction where hydrogen gas reacts with oxygen gas to form water in liquid form:

Given: - 2.0 L of hydrogen gas at STP reacts with excess oxygen to form liquid water.

Step-by-Step Solution:

• Convert volume of H₂ gas to moles. At STP, 1 mole of gas occupies 22.4 L.
• Moles of H₂ = 2.0 L / 22.4 L/mol ≈ 0.0893 mol
• Using the balanced equation: 2 H₂ + O₂ → 2 H₂O, 2 moles of H₂ produce 2 moles of H₂O.
• Therefore, 0.0893 mol of H₂ will produce 0.0893 mol of H₂O.
• Convert this to mass using the molar mass of H₂O, which is about 18.02 g/mol:
  • Mass of H₂O = 0.0893 mol * 18.02 g/mol ≈ 1.61 g

💦 Note: Water molecules in liquid form occupy a smaller volume than in gas phase, which is not calculated here as the problem specifies liquid water.

Problem 5: Solid to Gas to Solid Reaction

Finally, explore a complex scenario where calcium carbonate, heated, decomposes into calcium oxide and carbon dioxide, which can then react with water to form calcium hydroxide.

Given: - 10.0 grams of calcium carbonate (CaCO₃) are heated to produce CO₂ at STP, which then reacts with an excess of water to form calcium hydroxide (Ca(OH)₂).

Step-by-Step Solution:

• Calculate the moles of CaCO₃. The molar mass of CaCO₃ is approximately 100.09 g/mol.
• Moles of CaCO₃ = 10.0 g / 100.09 g/mol ≈ 0.0999 mol
• Using the balanced equation: CaCO₃(s) → CaO(s) + CO₂(g), 1 mole of CaCO₃ decomposes to produce 1 mole of CO₂.
• Therefore, 0.0999 mol of CO₂ is produced.
• CO₂(g) + H₂O(l) → Ca(OH)₂(s); here, 1 mole of CO₂ reacts to form 1 mole of Ca(OH)₂.
• Convert this to mass using the molar mass of Ca(OH)₂, which is about 74.093 g/mol:
  • Mass of Ca(OH)₂ = 0.0999 mol * 74.093 g/mol ≈ 7.406 g

🔥 Note: This reaction involves heating, which alters the state of matter and the reaction kinetics.

The exploration of mixed stoichiometry problems underscores the versatility and necessity of understanding how different states of matter interact in chemical reactions. From gas to gas, liquid to solid, solids to gas, gas to liquid, to solids to gas to solids, each reaction presents unique challenges that require a solid foundation in stoichiometry principles. These problems teach us how to apply concepts like molar volume, balancing equations, and using Avogadro's number effectively, providing a practical application of theoretical knowledge. Whether you're preparing for an exam, working in a lab, or simply curious about the chemistry around you, these examples serve as a comprehensive guide to navigate the complexities of chemical reactions in various conditions.