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Chapter 5: Problem 137

A chemist weighed out \(5.14 \mathrm{~g}\) of a mixture containing unknown amounts of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) and placed the sample in a \(1.50-\mathrm{L}\) flask containing \(\mathrm{CO}_{2}(g)\) at \(30.0^{\circ} \mathrm{C}\) and 750 . torr. After the reaction to form \(\mathrm{BaCO}_{3}(s)\) and \(\mathrm{CaCO}_{3}(s)\) was completed, the pressure of \(\mathrm{CO}_{2}(g)\) remaining was 230 . torr. Calculate the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture.

Short Answer

Expert verified
The mass percentages of BaO and CaO in the mixture are approximately 84.53% and 15.47%, respectively.

Step by step solution

01

Calculate moles of CO2 reacted

Subtract the final pressure of CO2 from the initial pressure to get the total pressure change of CO2 due to the reaction. - Initial CO2 pressure: 750 torr - Final CO2 pressure: 230 torr Total pressure change of CO2: 750 torr - 230 torr = 520 torr Using the ideal gas law (\(PV = nRT\)), we can find the moles of CO2 that reacted. Convert the pressure change to atmospheres, and use R = 0.0821 L atm/(mol K). - Pressure change in atm: 520 torr * (1 atm / 760 torr) = 0.684 atm - Volume: 1.50 L - Temperature: 30.0°C + 273.15 = 303.15 K \[n = \frac{P \cdot V}{R \cdot T} = \frac{0.684 \, atm \cdot 1.50 \, L}{0.0821 \frac{L \, atm}{mol \, K} \cdot 303.15 \, K} = 0.04254 \, mol\]
02

Use stoichiometry to find moles of BaO and CaO

Since BaO and CaO both react with CO2 in a 1:1 ratio, we can say that: - moles of BaO reacted + moles of CaO reacted = moles of CO2 reacted - Let x be the moles of BaO reacted, so (0.04254 - x) represents the moles of CaO reacted.
03

Calculate the mass of BaO and CaO

Use the molar mass of BaO and CaO to find their masses. The sum of their masses is equal to the given mixture mass (5.14 g). - Molar mass of BaO: 137.3 g/mol (Ba) + 16.0 g/mol (O) = 153.3 g/mol - Molar mass of CaO: 40.1 g/mol (Ca) + 16.0 g/mol (O) = 56.1 g/mol - Mass of BaO = x * molar mass of BaO = 153.3x g - Mass of CaO = (0.04254 - x) * molar mass of CaO = 56.1(0.04254 - x) g Sum of masses = 5.14 g, therefore: 153.3x + 56.1(0.04254 - x) = 5.14
04

Solve for x

Now, we'll solve the equation 153.3x + 56.1(0.04254 - x) = 5.14 for x: 153.3x + 56.1(0.04254 - x) = 5.14 153.3x + 2.383 - 56.1x = 5.14 97.2x = 2.757 x = 0.02835
05

Calculate mass percentages

Now that we have x, which represents the moles of BaO that reacted, we can find the mass percentages of BaO and CaO: - Mass of BaO = 153.3 * 0.02835 = 4.345 g - Mass of CaO = 5.14 g - 4.345 g = 0.795 g Mass percentage of BaO: \(\frac{4.345 \, g}{5.14 \, g} \times 100\% = 84.53\%\) Mass percentage of CaO: \(\frac{0.795 \, g}{5.14 \, g} \times 100\% = 15.47\%\) The mass percentages of BaO and CaO in the mixture are approximately 84.53% and 15.47%, respectively.

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Most popular questions from this chapter

Some very effective rocket fuels are composed of lightweight liquids. The fuel composed of dimethylhydrazine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{H}_{2}\right]\) mixed with dinitrogen tetroxide was used to power the Lunar Lander in its missions to the moon. The two components react according to the following equation: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{H}_{2}(l)+2 \mathrm{~N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)\) If \(150 \mathrm{~g}\) dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at \(27^{\circ} \mathrm{C}\) in an evacuated 250-L tank, what is the partial pressure of nitrogen gas produced and what is the total pressure in the tank assuming the reaction has \(100 \%\) yield?

Calculate the pressure exerted by \(0.5000 \mathrm{~mol} \mathrm{~N}_{2}\) in a 1.0000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

In the presence of nitric acid, \(\mathrm{UO}^{2+}\) undergoes a redox process. It is converted to \(\mathrm{UO}_{2}{ }^{2+}\) and nitric oxide (NO) gas is produced according to the following unbalanced equation: \(\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2+}(a q)\) \(\mathrm{NO}(g)+\mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) If \(2.55 \times 10^{2} \mathrm{~mL} \mathrm{NO}(g)\) is isolated at \(29^{\circ} \mathrm{C}\) and \(1.5 \mathrm{~atm}\), what amount (moles) of \(\mathrm{UO}^{2+}\) was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)

Ethene is converted to ethane by the reaction $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \stackrel{\mathrm{Cindys}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) $$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at \(25.0 \mathrm{~atm}\) and \(300 .{ }^{\circ} \mathrm{C}\) with a flow rate of \(1000 .\) L/min. Hydrogen at \(25.0\) atm and \(300 .{ }^{\circ} \mathrm{C}\) flows into the reactor at a flow rate of \(1500 . \mathrm{L} / \mathrm{min}\). If \(15.0 \mathrm{~kg} \mathrm{C}_{2} \mathrm{H}_{6}\) is collected per minute, what is the percent yield of the reaction?

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