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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 between hydrochloric acid and magnesium carbonate, and hydrochloric acid and calcium carbonate are: \[MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + CO_2(g) + H_2O(l)\] \[CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)\] Using the Ideal Gas Law, we find that the total moles of carbon dioxide formed is \(0.0732\,\text{mol}\). By setting up a mass balance equation and solving for the mass of magnesium carbonate, we find that it is \(3.67\, \text{g}\). Therefore, the percentage by mass of magnesium carbonate in the mixture is \(56.2\%\).

Step by step solution

01

a) Balanced Chemical Equations

For the reaction of hydrochloric acid (HCl) with magnesium carbonate (MgCO3): \[MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + CO_2(g) + H_2O(l)\] For the reaction of hydrochloric acid (HCl) with calcium carbonate (CaCO3): \[CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)\]
02

b) Calculate Total Moles of Carbon Dioxide

We can use the Ideal Gas Law formula, which is: \(PV = nRT\) Where: - P = pressure (in atm) - V = volume (in L) - n = moles (in mol) - R = ideal gas constant (0.0821 Lâ‹…atm/molâ‹…K) - T = temperature (in K) First, we convert the given pressure from torr to atm: \[743 \,\text{torr} \times \frac{1 \,\text{atm}}{760\, \text{torr}} = 0.976 \,\text{atm}\] Next, we convert the given temperature from Celsius to Kelvin: \[(28^\circ \text{C}) + 273.15 = 301.15 \,\text{K}\] Now, we can use the Ideal Gas Law formula to find the total moles (n) of carbon dioxide (CO2) formed: \[n_\text{CO2} = \frac{PV}{RT} = \frac{(0.976 \,\text{atm})(1.72\, \text{L})}{(0.0821 \, \text{Lâ‹…atm/molâ‹…K})(301.15 \,\text{K})} = 0.0732\, \text{mol}\]
03

c) Calculate Percentage by Mass of Magnesium Carbonate

Let x be the mass of magnesium carbonate (MgCO3) and (6.53 - x) be the mass of calcium carbonate (CaCO3) in the mixture. Then, the moles of CO2 produced from each component can be calculated as: For MgCO3: \(n_\text{CO2(MgCO3)}=\frac{x}{M_\text{MgCO3}}\) For CaCO3: \(n_\text{CO2(CaCO3)}=\frac{6.53-x}{M_\text{CaCO3}}\) The total moles of CO2 formed from both components is 0.0732 mol (from part b), so: \(\frac{x}{M_\text{MgCO3}} + \frac{6.53 - x}{M_\text{CaCO3}} = 0.0732\) Where: - \(M_\text{MgCO3}\) is the molar mass of magnesium carbonate, which is 84.31 g/mol. - \(M_\text{CaCO3}\) is the molar mass of calcium carbonate, which is 100.09 g/mol. Solve the equation for x: \(x = 3.67 \,\text{g}\) Finally, we can calculate the percentage by mass of magnesium carbonate in the mixture: \[\text{Percentage of MgCO3} = \frac{3.67 \,\text{g}}{6.53 \,\text{g}} \times 100\% = 56.2\%\]

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Key Concepts

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

Balanced chemical equations
In the context of chemical reactions, a balanced chemical equation is essential. It shows the relationship between the reactants and products in a chemical process. When balancing chemical equations, you ensure that the number of atoms of each element is the same on both sides of the equation. This is important because of the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a closed system.

For example, when hydrochloric acid (HCl) reacts with either magnesium carbonate (MgCO_3) or calcium carbonate (CaCO_3), it produces magnesium chloride (MgCl_2) or calcium chloride (CaCl_2), along with carbon dioxide (CO_2) and water (H_2O).
  • For the reaction between HCl and MgCO_3: \[ MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + CO_2(g) + H_2O(l) \]

  • For the reaction between HCl and CaCO_3: \[ CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l) \]

Balancing these equations involves ensuring that there are equal numbers of each type of atom on both sides of the equation. This way, the calculated results reflect the true chemical behavior in the reaction.
Ideal Gas Law
The Ideal Gas Law is a crucial tool in chemistry for relating the physical properties of gases. It combines several gas laws into a single equation to describe the state of an ideal gas:

\[ PV = nRT \]

Where:
  • P is the pressure of the gas (in atmospheres).
  • V is the volume of the gas (in liters).
  • n is the number of moles of gas.
  • R is the ideal gas constant (0.0821 Lâ‹…atm/molâ‹…K).
  • T is the temperature of the gas (in Kelvin).

To apply the Ideal Gas Law, one needs to ensure that all quantities are in the appropriate units. For instance, temperature must be converted to Kelvin by adding 273.15 to the Celsius measurement. Pressure must often be converted from units like Torr to atmospheres (using a conversion factor of 1 atm = 760 Torr).

By using the Ideal Gas Law, the number of moles of carbon dioxide produced can be calculated accurately from volume, temperature, and pressure data. In the given exercise, the calculation involves converting 1.72 L of CO_2 at 28°C and 743 torr into the number of moles of CO_2, providing vital information needed for further stoichiometric calculations.
Percentage composition by mass
Percentage composition by mass is a way of expressing the concentration of an element or compound in a mixture. It is calculated by taking the mass of the individual component, divided by the total mass of the mixture, and then multiplying by 100% to get the percentage.

In the context of the given problem, the percentage composition by mass of magnesium carbonate in the sample is determined after establishing the number of moles of CO_2 produced. For each component (MgCO_3 and CaCO_3), the number of moles of CO_2 they produce can be calculated using their respective balanced reactions and molar masses.

Once the moles are known, you can calculate the mass of each carbonate, and hence, use the formula:

\[ \text{Percentage by mass of component} = \left( \frac{\text{mass of component}}{\text{total mass of mixture}} \right) \times 100 \% \]

This calculation is crucial, as it gives insight into the composition of the sample, which is particularly useful in analytical chemistry for understanding the make-up of complex mixtures. In this case, solving the relevant equation for the mass of MgCO_3 provides its percentage in the original sample, resulting in a calculated percentage of 56.2%.

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Most popular questions from this chapter

Consider the combustion reaction between \(25.0 \mathrm{~mL}\) of liquid methanol (density \(=0.850 \mathrm{~g} / \mathrm{mL})\) and \(12.5 \mathrm{~L}\) of oxygen \(\mathrm{gas}\) measured at STP. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the volume of liquid \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion and you condense the water vapor.

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below \(100^{\circ} \mathrm{C}\) in a boiling-water bath and determine the mass of vapor required to fill the bulb (see drawing, next page). From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3} ;\) pressure, 742 torr; temperature, \(99^{\circ} \mathrm{C}\).

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations. Consider a 1000 -megawatt coalfired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, \(1.00 \mathrm{~atm}\), and \(27^{\circ} \mathrm{C},\) calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and \(120 \mathrm{~atm}\) and a density of \(1.2 \mathrm{~g} / \mathrm{cm}^{3},\) what volume does it possess? (c) If it is stored underground as a gas at \(36^{\circ} \mathrm{C}\) and \(90 \mathrm{~atm},\) what volume does it occupy?

The temperature of a 5.00-L container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root-mean-square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls ner second.

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 105 s for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ;\) that is, rate and time are inversely proportional.)

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