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 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

The balanced chemical equations for the reactions between hydrochloric acid and magnesium carbonate, and calcium carbonate are: \(MgCO_3 + 2HCl ⟶ MgCl_2 + H_2O + CO_2\) \(CaCO_3 + 2HCl ⟶ CaCl_2 + H_2O + CO_2\) Using the given conditions and the ideal gas law, we find that there are approximately 0.0732 moles of \(CO_2\) produced. Let x be the moles of \(MgCO_3\), so (0.0732 - x) are the moles of \(CaCO_3\). After solving for x, we find that there are approximately 0.0597 moles of \(MgCO_3\), corresponding to a mass of 5.03 g. The percentage by mass of magnesium carbonate in the mixture is approximately 77.1%.

Step-by-step solution

step 1: Balanced chemical equations

For magnesium carbonate and calcium carbonate reacting with hydrochloric acid, the unbalanced reactions are: \(MgCO_3 + HCl ⟶ MgCl_2 + H_2O + CO_2\) \(CaCO_3 + HCl ⟶ CaCl_2 + H_2O + CO_2\) Now we balance the reactions: \(MgCO_3 + 2HCl ⟶ MgCl_2 + H_2O + CO_2\) \(CaCO_3 + 2HCl ⟶ CaCl_2 + H_2O + CO_2\)

step 2: Use ideal gas law to find moles of \(CO_2\)

We are given the volume (1.72 L), temperature (28°C), and pressure (743 torr) of the carbon dioxide gas produced. First, we need to convert the temperature to Kelvin: \(T(K) = 28 + 273.15 = 301.15 K\) Next, convert pressure from torr to atm: \(P(atm) = 743 \frac{torr}{760 \frac{torr}{atm}} \approx 0.9776 atm\) Now use the ideal gas law formula to find the moles of \(CO_2\): \(PV=nRT\) where P = pressure (atm), V = volume (L), n = number of moles, R = gas constant = 0.0821 (L atm mol^(-1) K^(-1)), T = temperature (K) \(n = \frac{PV}{RT}\) Solving for n: \(n = \frac{(0.9776)(1.72)}{(0.0821)(301.15)}\) \(n ≈ 0.0732\) moles of \(CO_2\)

step 3: Calculate the percentage of MgCO3 in the mixture

Let x be the moles of \(MgCO_3\), so (0.0732 - x) are the moles of \(CaCO_3\). Based on the stoichiometry of the balanced chemical equations, one mole of each carbonate produces one mole of \(CO_2\). Hence, the masses of the carbonates in the mixture are: -Mass of \(MgCO_3 = x × M_{MgCO_3}\), where \(M_{MgCO_3} = 84.31 g/mol\) -Mass of \(CaCO_3 = (0.0732 - x) × M_{CaCO_3}\), where \(M_{CaCO_3} = 100.09 g/mol\) The mass of the mixture is given as 6.53 g. Therefore: \(84.31x + 100.09(0.0732 - x) = 6.53\) Solving this equation for x: \(x ≈ 0.0597\) moles of \(MgCO_3\) Now compute the mass of magnesium carbonate (MgCO3) and then find the percentage of MgCO3 in the mixture. Mass of \(MgCO_3 = 0.0597 × 84.31 = 5.03 g\) Percentage of \(MgCO_3 = \frac{5.03}{6.53} × 100 = 77.1\%\) The percentage by mass of magnesium carbonate in the mixture is approximately 77.1%.

Key concepts

Moles of Gas

When it comes to gases, understanding how many moles you have can be crucial. A mole is a way to count particles, much like a dozen counts items by twelves. The number of moles can be found from various gas properties using the Ideal Gas Law. The problem provides volume (1.72 L), pressure (743 torr), and temperature (28°C) for carbon dioxide (CO₂). Before calculating moles, we need to get those units right. Convert the temperature to Kelvin by adding 273.15 to Celsius so we get 301.15 K. Change pressure from torr to atm, since 1 atm equals 760 torr. Thus, 743 torr turns into approximately 0.9776 atm. Utilizing the Ideal Gas Law, you can solve for moles (## Ideal Gas LawUse \( PV = nRT \) to solve for \( n \):\[ n = \frac{PV}{RT} \]Where:- \( P \) is pressure in atm,- \( V \) is volume in liters,- \( R \) is the constant (0.0821 L atm mol⁻¹ K⁻¹), - \( T \) is temperature in Kelvin.In this calculation: \( n = \frac{(0.9776)(1.72)}{(0.0821)(301.15)} \approx 0.0732 \) moles.So, you have around 0.0732 moles of CO₂ produced from these reactions.

Chemical Equations

Chemical equations detail the substances before and after a reaction. They must be balanced, meaning the amount of each element must be equal on both sides. This ensures mass is conserved.For the given chemical reactions:- **Magnesium Carbonate:** - The unbalanced equation is: \( MgCO_3 + HCl \rightarrow MgCl_2 + H_2O + CO_2 \) - Balanced becomes: \( MgCO_3 + 2HCl \rightarrow MgCl_2 + H_2O + CO_2 \)- **Calcium Carbonate:** - Unbalanced: \( CaCO_3 + HCl \rightarrow CaCl_2 + H_2O + CO_2 \) - Balanced turns to: \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \)## Understanding BalancingEach reactant's atoms appear on both sides of the equations. We use coefficients, like '2' before HCl, to balance the number of atoms, confirming that neither matter nor energy is lost.

Ideal Gas Law

The Ideal Gas Law is a quintessential tool in chemistry, helping us relate different states of gas to its volume, amount, temperature, and pressure. It assumes gases behave ideally, meaning individual gas particles are point-like with no volume and no intermolecular forces. ## Formula Details- The law is shown as: \[ PV = nRT \] where: - \( P \) = pressure, - \( V \) = volume, - \( n \) = moles, - \( R \) = ideal gas constant (approximately 0.0821 L atm mol⁻¹ K⁻¹), - \( T \) = temperature (in Kelvin).Plug in values for any three to find the fourth. For instance, given \( P \), \( V \), and \( T \), you can solve for \( n \), the number of moles, which tells you the quantity of gas available.## Application InsightConsider the scenario with CO₂ gas: Knowing its pressure, volume, and temperature, you used the Ideal Gas Law to determine you had around 0.0732 moles of CO₂. This is crucial when you need to find out how much of a substance reacted to form another.

Mass Percentage

Mass percentage in a mixture indicates how much of one component exists compared to the total mass. It's a favorite in chemistry because it gives a clear, understandable comparison and accounts for all mixtures' parts.## Calculation in Steps1. **Determine Moles:** - Given total moles of CO₂ from the reaction was 0.0732. 2. **Link Moles to Reactants:** - Both MgCO₃ and CaCO₃ produce CO₂ in equal moles. - Let \( x \) be moles of MgCO₃; then \( (0.0732 - x) \) is for CaCO₃.3. **Mass Calculations:** - Mass of MgCO₃: \( x \times 84.31 \text{ g/mol} \) - Mass of CaCO₃: \( (0.0732 - x) \times 100.09 \text{ g/mol} \)4. **Mass Equation:** - Combine for total mass to solve: \[ 84.31x + 100.09(0.0732 - x) = 6.53 \text{ g} \]5. **MgCO₃ Mass Finding:** - Solve the above equation to find \( x \approx 0.0597 \). - Mass of MgCO₃ equals the moles times molar mass: \( 5.03 \text{ g} \)6. **Percentage Calculation:** - \( \frac{5.03}{6.53} \times 100 = 77.1\% \).This percentage reveals the dominance of magnesium carbonate in the mixture.