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Chapter 10: Problem 127

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

Expert verified
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

01

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: \[ \text{MgCO}_{3 \:(s)} + 2 \text{HCl}_{\:(aq)} \rightarrow \text{MgCl}_{2 \:(aq)} + \text{H}_{2 \text{O}}_{\:(l)} + \text{CO}_{2 \:(g)} \]
02

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#
03

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)
04

Solving for n

Solving for n, we get: \[ n = \dfrac{PV}{RT} \] \[ n = \dfrac{(99.06 \, \text{kPA})(1.72 \, \text{L})}{(8.314 \, \text{kPA L/mol K})(301.15 \, \text{K})} \]
05

Calculating n

Calculate the number of moles: \[ n = 0.085 \, \text{mol} \] #c) Percentage by mass of magnesium carbonate#
06

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.
07

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.
08

Equation relating the mass of the mixture

Write the equation: \[ x(24.31 + 12.01 + 3\times16) + (0.085 - x)(40.08 + 12.01 + 3\times16) = 6.53 \, \text{g} \]
09

Solving for x

Now, solve for x: \[ x = 0.045 \, \text{mol} \]
10

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%.

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Most popular questions from this chapter

A sample of \(5.00 \mathrm{~mL}\) of diethylether $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\( density \)=0.7134 \mathrm{~g} / \mathrm{mL}\( ) is introduced into a \)6.00-\mathrm{L}$ vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(R_{\mathrm{N}_{2}}=21.08 \mathrm{kPa}\) and \(P_{\mathrm{O}_{2}}=76.1 \mathrm{kPa}\). The temperature is held at \(35.0^{\circ} \mathrm{C},\) and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of $5.67 \mathrm{~L}\( has a partial pressure of \)\mathrm{O}_{2}\( of \)7.066 \mathrm{mPa}\( at \)30^{\circ} \mathrm{C}$, what mass of magnesium will react according to the following equation? $$2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s).$$

A mixture containing $0.50 \mathrm{~mol} \mathrm{H}_{2}(g), 1.00 \mathrm{~mol} \mathrm{O}_{2}(g)\(, and 3.50 \)\mathrm{mol} \mathrm{N}_{2}(g)$ is confined in a 25.0-L vessel at \(25^{\circ} \mathrm{C}\). (a) Calculate the total pressure of the mixture. (b) Calculate the partial pressure of each of the gases in the mixture.

(a) Calculate the density of dinitrogen tetroxide gas $\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\( at \)111.5 \mathrm{kPa}\( and \)0{ }^{\circ} \mathrm{C}$. (b) Calculate the molar mass of a gas if 2.70 g occupies \(0.97 \mathrm{~L}\) at \(134.7 \mathrm{~Pa}\) and \(100^{\circ} \mathrm{C}\).

Calculate the pressure that \(\mathrm{CCl}_{4}\) will exert at $80^{\circ} \mathrm{C}\( if 1.00 mol occupies \)33.3 \mathrm{~L}$, assuming that (a) \(\mathrm{CCl}_{4}\) obeys the ideal-gas equation; (b) \(\mathrm{CCl}_{4}\) obeys the van der Waals equation. (Values for the van der Waals constants are given in Table \(10.3 .\) ) (c) Which would you expect to deviate more from ideal behavior under these conditions, \(\mathrm{Cl}_{2}\) or \(\mathrm{CCl}_{4} ?\) Explain.
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