Problem 153 - Chemistry

Question

A mixture of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) and magnesium carbonate \(\left(\mathrm{MgCO}_{3}\right)\) of mass \(6.26 \mathrm{~g}\) reacts completely with hydrochloric acid (HCl) to generate \(1.73 \mathrm{~L}\) of \(\mathrm{CO}_{2}\) at \(48^{\circ} \mathrm{C}\) and 1.12 atm. Calculate the mass percentages of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) in the mixture.

Step-by-Step Solution

Answer

The mass percentage of \(\mathrm{CaCO}_{3}\) in the mixture is 63.1%, and the mass percentage of \(\mathrm{MgCO}_{3}\) in the mixture is 36.9%.

Step 1: Identify the reactions

The reactions that take place are: \(\mathrm{CaCO}_{3}(s) + 2\mathrm{HCl}(aq) → \mathrm{CaCl}_{2}(aq) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\) \(\mathrm{MgCO}_{3}(s) + 2\mathrm{HCl}(aq) → \mathrm{MgCl}_{2}(aq) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\) In these reactions, 1 mole of each type of carbonate gives 1 mole of CO2. This fact will be important when accounting for gas produced.

Step 2: Determine moles of CO2 produced

Use the ideal gas equation: \(PV=nRT\) changing given conditions to required units: \(P=1.12\,atm\), \(V=1.73\,L\), \(T=48+273.15=321.15\,K\) and \(R=0.0821\,atm.L/mol.K\), solve for \(n=(PV)/(RT)=((1.12*1.73)/(0.0821*321.15))=0.073\,moles\). There are 0.073 moles of CO2 produced.

Step 3: Calculate theoretical moles of CO2 each compound could produce

The potential moles of CO2 each compound can produce is equal to the moles of the compound present. Hence, the moles of CaCO3 = moles of CO2 it can produce and moles of MgCO3 = moles of CO2 it can produce. Let's denote the mass (in g) of CaCO3 as \(x\) and of MgCO3 as \(6.26-x\). Hence, the moles of CaCO3 present = \(x/100.09\) and the moles of MgCO3 present = \((6.26-x)/84.31\), as the molar mass of CaCO3 is 100.09 g/mol and that of MgCO3 is 84.31 g/mol. Thus, the amount of CO2 that each compound can produce can be expressed in terms of \(x\).

Step 4: Equate total CO2 to CO2 produced by individual compounds

Since given total moles = sum of moles from each compound, we can write: 0.073 \(moles\) = \(x/100.09 + (6.26-x)/84.31\), which allows us to solve for \(x = 3.95\,g\).

Step 5: Calculate mass percentages

Percent mass of calcium carbonate = \((x/6.26)*100 = 63.1%\). Percent mass of magnesium carbonate = \(100 - 63.1% = 36.9%\)

Key Concepts

Ideal Gas Equation

The ideal gas equation is a fundamental equation used in chemistry to relate the four essential properties of gases: pressure, volume, temperature, and number of moles. It is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas in atm
• \( V \) is the volume of the gas in liters
• \( n \) is the number of moles of the gas
• \( R \) is the ideal gas constant \( 0.0821 \, \text{L atm} / \text{mol K} \)
• \( T \) is the temperature in Kelvin

To use this equation, you must ensure that the units are consistent. In our example with the calcium carbonate and magnesium carbonate mixture, we used the ideal gas equation to find the number of moles of the produced \( \text{CO}_2 \) gas. Remembering to convert temperature to Kelvin by adding 273.15 to the Celsius value is crucial. This formula allows us to calculate any one of these properties, provided the other three are known.

Calcium Carbonate

Calcium carbonate, \( \text{CaCO}_3 \), is a common chemical substance found in rocks as the minerals calcite and aragonite, most notably as limestone, which is the core component of eggshells, snails, shells, and pearls. In the context of chemical reactions, calcium carbonate reacts with acids, such as hydrochloric acid \( \text{HCl} \), to produce carbon dioxide gas, water, and a salt. The particular reaction in this exercise is:

• \( \text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \)

This is an example of an acid-carbonate reaction, where the carbonate compound releases \( \text{CO}_2 \) gas. Understanding this type of chemical reaction is important in the study of basic chemistry and geological formations.

Magnesium Carbonate

Magnesium carbonate, \( \text{MgCO}_3 \), is another naturally occurring mineral that is often used in tablets as an antacid and a laxative. Similar to calcium carbonate, it undergoes a reaction with hydrochloric acid to form magnesium chloride, carbon dioxide, and water:

• \( \text{MgCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l) \)

In this exercise, magnesium carbonate provides an additional source of \( \text{CO}_2 \) gas. Knowing its molar mass (84.31 g/mol) helps in calculating its contribution to the total gas produced, which is essential when determining the mass percentages within a mixture.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They can be categorized into many types, including synthesis, decomposition, single replacement, and double replacement reactions. In this exercise, we focus on reactions between carbonates and acids, which are a form of double replacement reaction.

• When carbonates react with acids like hydrochloric acid, they produce carbon dioxide gas, water, and a new salt compound.
• These reactions are typically exothermic, releasing some heat.

For calculating reactants and products, stoichiometry is used. It involves using the coefficients from balanced chemical equations to determine the relative amounts. Understanding these reaction principles allows for accurate predictions of outcome quantities.