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Chapter 8: Problem 100

The oxides of Group \(2 \mathrm{A}\) metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$\mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s)$$ A \(2.85\) \(-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00 -L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of \(740 .\) torr at \(20 .^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is \(390 .\) torr at \(20 .^{\circ} \mathrm{C}\). What is the mass percent of \(MgO\) in the mixture? Assume that only the \(MgO\) reacts with \(\mathrm{CO}_{2}\)

Short Answer

Expert verified
The mass percent of MgO in the mixture is approximately 78.2 %.

Step by step solution

01

Calculate moles of CO2 initially present

We are given the initial pressure of CO2, the volume of the container, and the temperature. Using the Ideal Gas Law, we can calculate the moles of CO2 initially present: \(PV = nRT\) Where, P = Pressure in atm = 740 Torr × (1 atm / 760 Torr) = 0.9737 atm V = Volume = 3.00 L R = Gas constant = 0.0821 L atm/mol K T = Temperature in Kelvin = 20°C + 273.15 = 293.15 K \[n_{CO_2\, initial} = \frac{PV}{RT} = \frac{(0.9737)(3.00)}{(0.0821)(293.15)} \approx 0.1190 \,mol\]
02

Calculate moles of CO2 remaining after the reaction

Similarly, we can calculate the moles of CO2 remaining after the reaction using the Ideal Gas Law and the final conditions: P( final) = 390 Torr × (1 atm / 760 Torr) = 0.5132 atm \[n_{CO_2\, final} = \frac{PV}{RT} = \frac{(0.5132)(3.00)}{(0.0821)(293.15)} \approx 0.0637 \,mol\]
03

Calculate moles of CO2 reacted with MgO

The difference between the initial and final moles of CO2 is equal to the moles of CO2 that reacted with MgO: \[n_{CO_2\, reacted} = n_{CO_2\, initial} - n_{CO_2\, final} = 0.1190 - 0.0637 = 0.0553 \, mol\]
04

Determine moles of MgO

From the balanced reaction, we know the stoichiometry of CO2 and MgO is 1:1. Therefore, the moles of MgO must be equal to the moles of CO2 reacted: \[n_{MgO} = n_{CO_2\, reacted} = 0.0553 \,mol\]
05

Calculate mass of MgO in the mixture

To calculate the mass of MgO in the mixture, multiply the number of moles of MgO by its molar mass: Molar mass of MgO = 24.31 g/mol (Mg) + 16.00 g/mol (O) = 40.31 g/mol Mass of MgO = n_{MgO} × Molar mass of MgO = 0.0553 mol × 40.31 g/mol ≈ 2.23 g
06

Calculate the mass percent of MgO in the mixture

Now that we know the mass of MgO in the mixture, we can find the mass percent by dividing the mass of MgO by the total mass of the mixture (2.85 g) and multiplying by 100%: Mass percent of MgO = (Mass of MgO / Total mass) × 100 % = (2.23 g / 2.85 g) × 100 % ≈ 78.2 % The mass percent of MgO in the mixture is approximately 78.2 %.

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Most popular questions from this chapter

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) .\) Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is \(0.79 \mathrm{g} / \mathrm{cm}^{3} .\) )

Consider the following chemical equation. $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ If \(25.0 \mathrm{mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

A steel cylinder contains \(5.00\) moles of graphite (pure carbon) and \(5.00\) moles of \(\mathrm{O}_{2}\). The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by \(17.0 \% .\) Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 L. If the balloon is filled with 2.0 L helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only \(500 . \mathrm{mm}\) Hg, will the balloon burst? (Assume temperature is constant.)

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g),\) ammonia, \(\mathrm{NH}_{3}(g),\) and oxygen, \(\mathrm{O}_{2}(g),\) at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of HCN( \(g\) ) can be obtained from the reaction of \(20.0 \mathrm{L} \mathrm{CH}_{4}(g), 20.0 \mathrm{L} \mathrm{NH}_{3}(g),\) and \(20.0 \mathrm{LO}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.
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