Question

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 \(99.06 \mathrm{kPa}\) 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

The balanced chemical equations for the reactions between hydrochloric acid and magnesium carbonate, and hydrochloric acid and calcium carbonate are: \( \text{MgCO}_{3 \:(s)} + 2 \text{HCl}_{\:(aq)} \rightarrow \text{MgCl}_{2 \:(aq)} + \text{H}_{2 \text{O}}_{\:(l)} + \text{CO}_{2 \:(g)} \) \( \text{CaCO}_{3 \:(s)} + 2 \text{HCl}_{\:(aq)} \rightarrow \text{CaCl}_{2 \:(aq)} + \text{H}_{2 \text{O}}_{\:(l)} + \text{CO}_{2 \:(g)} \) The total number of moles of carbon dioxide that forms from these reactions is 0.085 moles. Assuming that the reactions are complete, the percentage by mass of magnesium carbonate in the mixture is approximately 41.27%.

Step-by-step solution

Equation for magnesium carbonate reaction with hydrochloric acid

The first step is writing the balanced chemical equation for the reaction between magnesium carbonate and hydrochloric acid: \[ \text{MgCO}_{3 \:(s)} + 2 \text{HCl}_{\:(aq)} \rightarrow \text{MgCl}_{2 \:(aq)} + \text{H}_{2 \text{O}}_{\:(l)} + \text{CO}_{2 \:(g)} \]

Equation for calcium carbonate reaction with hydrochloric acid

Next, write the balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid: \[ \text{CaCO}_{3 \:(s)} + 2 \text{HCl}_{\:(aq)} \rightarrow \text{CaCl}_{2 \:(aq)} + \text{H}_{2 \text{O}}_{\:(l)} + \text{CO}_{2 \:(g)} \] #b) Total number of moles of carbon dioxide#

Ideal gas law

To determine the total number of moles of carbon dioxide, we can use the Ideal Gas Law formula: \[ PV = nRT \] Where: - P = pressure (99.06 kPa) - V = volume (1.72 L) - n = number of moles of CO2 - R = Ideal Gas Constant (8.314 J/mol K, for this exercise we will use 8.314 kPa L/mol K for the units to match with the exercise) - T = temperature in Kelvin (28°C + 273.15 = 301.15 K)

Solving for n

Solving for n, we get: \[ n = \dfrac{PV}{RT} \] \[ n = \dfrac{(99.06 \, \text{kPA})(1.72 \, \text{L})}{(8.314 \, \text{kPA L/mol K})(301.15 \, \text{K})} \]

Calculating n

Calculate the number of moles: \[ n = 0.085 \, \text{mol} \] #c) Percentage by mass of magnesium carbonate#

Moles of magnesium carbonate and calcium carbonate

Since each mole of magnesium carbonate and calcium carbonate produces one mole of carbon dioxide, the sum of the moles of magnesium carbonate and calcium carbonate is equal to the moles of carbon dioxide: 0.085 moles. Let x moles be magnesium carbonate and (0.085 - x) moles be calcium carbonate.

Mass of magnesium carbonate and calcium carbonate

Calculate the mass of magnesium carbonate and calcium carbonate (by using their molar masses): - Mass of magnesium carbonate (MgCO3) = x * (24.31+12.01+3*16) g/mol - Mass of calcium carbonate (CaCO3) = (0.085 - x) * (40.08+12.01+3*16) g/mol The mass of the mixture is 6.53 g. So, the sum of the mass of magnesium carbonate and calcium carbonate is equal to 6.53 g.

Equation relating the mass of the mixture

Write the equation: \[ x(24.31 + 12.01 + 3\times16) + (0.085 - x)(40.08 + 12.01 + 3\times16) = 6.53 \, \text{g} \]

Solving for x

Now, solve for x: \[ x = 0.045 \, \text{mol} \]

Percentage of magnesium carbonate

Calculate the mass of magnesium carbonate and the percentage of magnesium carbonate: \) \text{Mass of MgCO}_{3\,} = (0.045\,\text{mol}) \times (24.31 + 12.01 + 3 \times 16 \,\text{g/mol}) \) \) \text{Mass of MgCO}_{3\,} = 2.692 \, \text{g} \) Percentage of magnesium carbonate in the mixture: \) \% \text{MgCO}_{3\,} = \dfrac{2.692 \, \text{g}}{6.53 \, \text{g}} \times 100 \) ) \% \text{MgCO}_{3\,} = 41.27 \% \) The percentage by mass of magnesium carbonate in the mixture is approximately 41.27%.

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into new substances, called products. In this exercise, magnesium carbonate (MgCO₃) and calcium carbonate (CaCO₃) react with hydrochloric acid (HCl) to produce carbon dioxide (CO₂), water (H₂O), and corresponding chlorides. Let's break this down further.

• For magnesium carbonate: \[ \text{MgCO}_{3 ext{(s)}} + 2 \text{HCl}_{\text{(aq)}} \rightarrow \text{MgCl}_{2\text{(aq)}} + \text{H}_{2\text{O}}_{\text{(l)}} + \text{CO}_{2\text{(g)}} \]
This equation shows that one mole of magnesium carbonate reacts with two moles of hydrochloric acid to produce one mole of each product.
• For calcium carbonate:\[ \text{CaCO}_{3 ext{(s)}} + 2 \text{HCl}_{\text{(aq)}} \rightarrow \text{CaCl}_{2\text{(aq)}} + \text{H}_{2\text{O}}_{\text{(l)}} + \text{CO}_{2\text{(g)}} \]
This is similar to the magnesium equation, where one mole of calcium carbonate also yields one mole of CO₂.

Understanding these balanced equations is essential because it helps us predict the amount of products formed, such as the volume of CO₂ in this exercise. It also ensures that the law of conservation of mass is maintained, as no atoms are lost or gained in the process.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle that relates the pressure, volume, temperature, and number of moles of a gas. The formula is expressed as:\[ PV = nRT \] Where:

• \( P \) is the pressure of the gas in kilopascals (kPa).
• \( V \) is the volume of the gas in liters (L).
• \( n \) is the number of moles of the gas.
• \( R \) is the ideal gas constant, which is 8.314 kPa L/mol K.
• \( T \) is the temperature in Kelvin (K). (Convert Celsius to Kelvin by adding 273.15.)

In the given exercise, we use the Ideal Gas Law to find the total moles of carbon dioxide produced:1. Convert the temperature from Celsius to Kelvin: \( 28^{\circ} \text{C} + 273.15 = 301.15 \text{ K} \)2. Use the formula to solve for \( n \): \[ n = \frac{PV}{RT} = \frac{(99.06 \, \text{kPa})(1.72 \, \text{L})}{(8.314 \, \text{kPa L/mol K})(301.15 \, \text{K})} \]3. This calculation shows the total moles of CO₂ to be approximately 0.085 mol, which helps us connect with the reactions to find out how much of each carbonate was used.

Mass Percentage Calculation

Mass percentage calculation is crucial in determining the composition of a mixture by expressing the mass of a component as a percentage of the total mass. To find the mass percentage of magnesium carbonate in this exercise, we need to determine how much of it is present in the mixture.
Here's how you can calculate it:

• From the stoichiometry of the reaction, each mole of MgCO₃ produces one mole of CO₂. If \( x \) is the moles of MgCO₃, and total moles of CO₂ produced is 0.085 mol, then for calcium carbonate, we have \( 0.085-x \) moles.
• Calculate the mass of magnesium carbonate: \[ x(24.31 + 12.01 + 3 \times 16) \text{ g/mol} \]
• Calculate the mass of calcium carbonate similarly.
• The sum of these two masses equals the mixture's mass (6.53 g). Solve for \( x \) to find the mass of magnesium carbonate: 2.692 g.
• Calculate the percentage: \[ \% \text{MgCO}_{3} = \frac{2.692 \, \text{g}}{6.53 \, \text{g}} \times 100 = 41.27\% \]

This means approximately 41.27% of the original sample is magnesium carbonate, helping us understand the composition within the mixture. With this percentage, you can determine the purity or specific content in similar chemical problems.