Problem 127 - Chemistry The Central Science In Si Units

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.

Step-by-Step Solution

Answer

The magnesium carbonate comprises 14.41% of the mixture by mass.

Step 1: Write Balanced Chemical Equations

First, we need to write the balanced chemical equations for the reactions of each carbonate with hydrochloric acid. 1. For magnesium carbonate: \[ \text{MgCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{MgCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \] 2. For calcium carbonate: \[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \] Both reactions produce carbon dioxide gas along with water and the respective chloride salt.

Step 2: Calculate Moles of Carbon Dioxide Formed

The volume of carbon dioxide gas is given as \(1.72\, \text{L}\), with the conditions set at \(28^{\circ}\, \text{C}\) and \(99.06\, \text{kPa}\).To find the number of moles, we can use the Ideal Gas Law:\[ PV = nRT \]Rearrange to solve for \(n\):\[ n = \frac{PV}{RT} \]Where:- \(P = 99.06\, \text{kPa} = 99.06\, \text{kPa} \times \frac{1000\, \text{Pa}}{1\, \text{kPa}} = 99060\, \text{Pa} \)- \(V = 1.72\, \text{L} = 0.00172\, \text{m}^3\)- \(R = 8.314\, \text{J/mol}\cdot\text{K} \)- \(T = 28^{\circ}\, \text{C} + 273.15 = 301.15\, \text{K} \)Plugging in the values, we get:\[ n = \frac{99060 \times 0.00172}{8.314 \times 301.15} = 0.0671\, \text{mol} \]So, the total moles of carbon dioxide produced is \(0.0671\, \text{mol} \).

Step 3: Determine Mass Percentage of Magnesium Carbonate

Let us denote the moles of \(\text{MgCO}_3\) as \(x\) and \(\text{CaCO}_3\) as \(y\). From the chemical equation, each mole of carbonate produces one mole of \(\text{CO}_2\). Therefore:\[ x + y = 0.0671 \]Given the molar masses \(\text{MgCO}_3 = 84.31\, \text{g/mol}\) and \(\text{CaCO}_3 = 100.09\, \text{g/mol}\), the mass equation becomes:\[ 84.31x + 100.09y = 6.53 \]Using the substitution from the \(x+y = 0.0671\) equation:\[ y = 0.0671 - x \]Substitute \(y\) in the mass equation:\[ 84.31x + 100.09(0.0671-x) = 6.53 \]Simplifying gives us:\[ 84.31x + 6.706 - 100.09x = 6.53 \]\[ -15.78x = -0.176 \]\[ x = 0.01116 \]The mass of \(\text{MgCO}_3\) is:\[ 0.01116 \times 84.31 = 0.941 \text{g} \]Calculate the percentage by mass of \(\text{MgCO}_3\):\[ \left(\frac{0.941}{6.53}\right) \times 100 = 14.41\% \]

Step 4: Conclusion

The magnesium carbonate makes up 14.41% of the original mixture by mass.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a crucial equation for understanding how gases behave under different conditions. It relates pressure, volume, and temperature of a gas to the number of moles present in the system. The equation is expressed as:
\[ PV = nRT \]where:

• P is the pressure of the gas in Pascals (Pa).
• V is the volume in cubic meters (m³).
• n is the number of moles of gas.
• R is the ideal gas constant, approximately eq to 8.314 J/(mol·K).
• T is the temperature in Kelvin (K).

In the given exercise, the conditions are specified as a temperature of 28°C and a pressure of 99.06 kPa. These need to be converted to Kelvin and Pascals, respectively, before using the Ideal Gas Law.
To find the number of moles of carbon dioxide produced, you can rearrange the Ideal Gas Law to solve for \( n \):
\[ n = \frac{PV}{RT} \]This allows calculation of how many moles are present based on the given conditions of the gas.

Moles Calculation

Calculating moles is fundamental to understanding chemical reactions. A mole is a unit that expresses amounts of a chemical substance. One mole contains exactly \( 6.022 \times 10^{23} \) (Avogadro's number) of molecules or atoms.
In this exercise, the calculation of moles is derived from the information obtained using the Ideal Gas Law. By rearranging the formula, the number of moles of carbon dioxide (\( CO_2 \)) produced is calculated.
The provided solution shows exact usage with conditions like:

• Pressure is converted from kPa to Pascals by multiplying by 1000.
• Temperature is converted from Celsius to Kelvin by adding 273.15.
• Volume of gas is adjusted to cubic meters.

After these conversions, you substitute the values into the Ideal Gas Equation to find \( n \), representing moles of \( CO_2 \) formed in the reaction. This foundational concept helps in predicting and understanding the extent of the reaction.

Chemical Equations Balancing

Balancing chemical equations is an integral part of chemistry. It ensures that the number of atoms for each element is conserved across a reaction, reflecting the law of conservation of mass.
In the exercise, the reactions of magnesium carbonate (\( ext{MgCO}_3 \)) and calcium carbonate (\( ext{CaCO}_3 \)) with hydrochloric acid (\( ext{HCl} \)) were balanced. These are typical reactions producing chloride salts, water, and carbon dioxide. The balanced equations are:
\[ \text{MgCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{MgCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \]
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g) \]Note how each carbonate reacts with two moles of hydrochloric acid to produce one mole of carbon dioxide. This balanced representation makes it easier to perform stoichiometric calculations, determining the molar relationships between reactants and products.

Mass Percentage Calculation

Calculating the mass percentage in a mixture is necessary to determine the composition and properties of solutions and compounds. To find the percentage by mass, you calculate the mass of the component of interest relative to the total mass of the mixture.
In this problem, the focus is on determining the mass percentage of magnesium carbonate (\( ext{MgCO}_3 \)) in a mixture containing calcium carbonate (\( ext{CaCO}_3 \)) as well. After calculating the moles of \( ext{MgCO}_3 \) using the molar mass, the mass is determined:
\[ \text{Mass} = \text{moles} \times \text{molar mass} \]The percent composition is then calculated by:
\[ \text{Percent by mass of } ext{MgCO}_3 = \left(\frac{\text{mass of } ext{MgCO}_3}{\text{total mass of mixture}}\right) \times 100 \]Applying these straightforward calculations will yield the proportion of magnesium carbonate present, expressed as a percentage.