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Chapter 10: Problem 122

A 6.53-g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochloric acid. The resulting reaction produces \(1.72 \mathrm{~L}\) of carbon dioxide gas at \(28^{\circ} \mathrm{C}\) and 743 torr pressure. (a) Write balanced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of carbon dioxide that forms from these reactions. (c) Assuming that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

Short Answer

Expert verified
The balanced chemical equations for the reactions are: \(MgCO_3 (s) + 2HCl (aq) \rightarrow MgCl_2 (aq) + H_2O (l) + CO_2 (g)\) \(CaCO_3 (s) + 2HCl (aq) \rightarrow CaCl_2 (aq) + H_2O (l) + CO_2 (g)\) Using the ideal gas law, we find that the total number of moles of CO₂ produced is 0.0679 mol. After setting up and solving a mass conservation equation for the mixture, we determine that there are 3.84 g of magnesium carbonate in the mixture. The mass percentage of magnesium carbonate in the mixture is approximately 58.8%.

Step by step solution

01

Balanced chemical equations for the reactions

First, let's write the balanced chemical equations for the reactions between hydrochloric acid and magnesium carbonate, and hydrochloric acid and calcium carbonate. For magnesium carbonate and hydrochloric acid: \(MgCO_3 (s) + 2HCl (aq) → MgCl_2 (aq) + H_2O (l) + CO_2 (g)\) For calcium carbonate and hydrochloric acid: \(CaCO_3 (s) + 2HCl (aq) → CaCl_2 (aq) + H_2O (l) + CO_2 (g)\) Both reactions produce one mole of CO2 for each mole of carbonate reacted.
02

Calculate total moles of carbon dioxide

We are given the following conditions for the carbon dioxide gas produced: volume = 1.72 L, temperature = 28°C (or 301.15 K), and pressure = 743 torr. We will use the ideal gas law to calculate the total number of moles of carbon dioxide. Ideal gas law: \(PV = nRT\) In this equation, P is pressure, V is volume, n is number of moles, R is the universal gas constant (\(R = 62.3637 \frac{\mathrm{L\cdot torr}}{\mathrm{mol\cdot K}}\)), and T is temperature. We can calculate the total number of moles of carbon dioxide, n, as follows: \(n = \frac{PV}{RT}\) \(n = \frac{(743\ \mathrm{torr})(1.72\ \mathrm{L})}{(62.3637\ \frac{\mathrm{L\cdot torr}}{\mathrm{mol\cdot K}})(301.15\ \mathrm{K})}\) \(n = 0.0679\ \mathrm{mol}\)
03

Mass conservation equation

Now, let's set up a mass conservation equation for the mixture of magnesium carbonate (MgCO3) and calcium carbonate (CaCO3). Let x be the mass of magnesium carbonate and y be the mass of calcium carbonate. Then, the mass of the mixture is given by: Total mass of magnesium carbonate + total mass of calcium carbonate = 6.53 g \(x + y = 6.53\ \mathrm{g}\) (1) Since one mole of magnesium carbonate and one mole of calcium carbonate both produce one mole of carbon dioxide, we can also write an equation for the moles of carbon dioxide produced from both these reactions: \(\frac{x}{\mathrm{MgCO_3}} + \frac{y}{\mathrm{CaCO_3}} = 0.0679\ \mathrm{mol}\) (2) Note that \(M_{\mathrm{MgCO_3}} = 84.31\ \mathrm{g/mol}\) and \(M_{\mathrm{CaCO_3}} = 100.09\ \mathrm{g/mol}\).
04

Calculate mass percentage of magnesium carbonate

We can now solve equation (2) for y in terms of x and substitute the result in equation (1) to solve for x: \(y = \frac{6.53 - x}{1}\) Substitute this into equation (2): \(\frac{x}{84.31} + \frac{6.53 - x}{100.09} = 0.0679\) Solve for x: \(x = 3.84\ \mathrm{g}\) of magnesium carbonate in the mixture. Now, we can calculate the percentage by mass of magnesium carbonate in the mixture: Mass percentage of magnesium carbonate = \(\frac{\text{mass of magnesium carbonate}}{\text{total mass of the mixture}}\) × 100% Mass percentage of magnesium carbonate = \(\frac{3.84\ \mathrm{g}}{6.53\ \mathrm{g}}\) × 100% Mass percentage of magnesium carbonate = 58.8% Therefore, the percentage by mass of magnesium carbonate in the mixture is approximately 58.8%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
In the context of solving stoichiometry problems, understanding chemical reactions is pivotal. A chemical reaction involves the transformation of reactants into products through the breaking and forming of chemical bonds. For students grappling with stoichiometry, it's essential to first write out the balanced chemical equations, as they clearly indicate the ratio of reactants to products. In our exercise example, the balanced equations for the reaction of magnesium carbonate and calcium carbonate with hydrochloric acid showed a one-to-one mole ratio between the carbonate compounds and the produced carbon dioxide gas. This balance is the foundation for all subsequent calculations in stoichiometry. It’s not just about recognizing chemical changes; it’s about quantifying those changes, which is where our next topic, the ideal gas law, comes into play.

For further clarity, remember that each reactant and product in a chemical equation is conserved. This is where we see the law of conservation of mass in action, which states that mass is neither created nor destroyed in a chemical reaction. With balanced equations, we ensure that the number of atoms for each element is the same on both sides.
Ideal Gas Law
The ideal gas law provides a clear relationship between the pressure, volume, temperature, and number of moles of a gas. When looking at a stoichiometry problem involving gases, the ideal gas law becomes an indispensable tool. In the example given, we applied the ideal gas law formula, \(PV = nRT\), to find the number of moles of carbon dioxide produced in the reactions. Understanding the conditions under which the gas exists (pressure, volume, and temperature) allows us to plug in values and solve for the unknowns. Keep in mind that R is the universal gas constant, and its value depends on the units used for pressure, temperature, and volume.

To assist students in mastering this concept, it's helpful to practice converting units (like torr to atm or Celsius to Kelvin) and using the appropriate R constant. When students comprehend this law, they can predict how gaseous reactants and products will behave under different conditions, which is a central aspect of chemical reactions and stoichiometry.
Mass Conservation
The mass conservation principle is fundamental to stoichiometry problems, which dictate that matter cannot be created or destroyed in a chemical reaction. This principle underlies the entire practice of stoichiometry—every calculation we perform is based on this conservation. In our exercise, we postulated that the mass of the mixture of magnesium carbonate and calcium carbonate equaled 6.53 g. This total mass is divided between the two compounds, and through stoichiometric calculations, we determined how much of each component made up the mixture.

To make these concepts clearer for students, it's beneficial to incorporate visual aids, like diagrams showing mass before and after a reaction or use algebraic methods as shown in the step-by-step solution to demonstrate how the masses of reactants correlate with the masses of products. Encouraging students to set up a mass balance table may also help them visually track where each component of the reaction is accounted for and understand the practical application of the mass conservation principle.

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Most popular questions from this chapter

A neon sign is made of glass tubing whose inside diameter is \(2.5 \mathrm{~cm}\) and whose length is \(5.5 \mathrm{~m}\). If the sign contains neon at a pressure of \(1.78\) torr at \(35^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h .\)

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