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Chapter 6: Problem 110

Aluminum (Al) and iron (Fe) each react with hydrochloric acid solution (HCl) to produce a chloride salt and hydrogen gas, \(\mathrm{H}_{2}(\mathrm{g}) .\) A \(0.1924 \mathrm{g}\) sample of a mixture of \(\mathrm{Al}\) and \(\mathrm{Fe}\) is treated with excess \(\mathrm{HCl}\) solution. A volume of \(159 \mathrm{mL}\) of \(\mathrm{H}_{2}\) gas is collected over water at \(19.0^{\circ} \mathrm{C}\) and 841 Torr. What is the percent (by mass) of Fe in the mixture? The vapor pressure of water at \(19.0^{\circ} \mathrm{C}\) is 16.5 Torr.

Short Answer

Expert verified
The percentage (by mass) of Fe in the mixture is 34.0 %.

Step by step solution

01

Calculate the pressure of Hydrogen Gas

The total pressure given is the sum of the pressure of hydrogen gas and the vapor pressure of water. So, the pressure of the hydrogen gas will be total pressure subtracted by the vapor pressure of water. Hence, \(P_{H_{2}} = P_{total} - P_{H_{2}O} = 841 Torr - 16.5 Torr = 824.5 Torr \)
02

Convert pressure to atm and volume to L

To apply ideal gas law, convert the pressure from Torr to atm by dividing by 760 (1 atm = 760 Torr), and the volume from mL to L by multiplying by 0.001 (1L = 1000mL). Thus, \[ P_{H_{2}} = \frac{824.5 Torr}{760 Torr/atm} = 1.085 atm \] and \[ V_{H_{2}} = 159 mL*0.001 L/mL = 0.159 L\]
03

Calculate the amount of Hydrogen Gas

Use the ideal gas law \( PV = nRT \) where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, T is the temperature. Solve for n (number of moles of hydrogen gas). \[ n_{H_{2}} = \frac{P_{H_{2}}V_{H_{2}}}{RT} = \frac{(1.085 atm)*(0.159 L)}{(0.0821 L*atm/mol*K)*(292 K)} = 0.00708 mol \]
04

Calculate the mass of Aluminum

From the balanced chemical reaction of Aluminum with hydrochloric acid, 3 moles of H2 are produced for each 2 moles of Aluminum. So, the moles of Aluminum reacting is \[ n_{Al} = \frac{2}{3}* n_{H_{2}} = \frac{2}{3}*0.00708 mol = 0.00472 mol \] Given Aluminum's molar mass is 26.98 g/mol, the mass of Aluminium can be calculated as \[ mass_{Al} = n_{Al} * molar mass_{Al} = 0.00472 mol * 26.98 g/mol = 0.127 g \]
05

Calculate the mass and percent of Iron

The mass of iron in the sample will be the total sample mass subtracted by the mass of aluminum. So, \[ mass_{Fe} = mass_{total} - mass_{Al} = 0.1924 g - 0.127 g = 0.0654 g \] The percentage of iron in the sample is calculated by \[ \% Fe = \frac{mass_{Fe}}{mass_{total}}* 100\% = \frac{0.0654 g}{0.1924 g}* 100\% = 34.0\% \]

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Most popular questions from this chapter

According to the CRC Handbook of Chemistry and Physics (83rd ed.), the molar volume of \(\mathrm{O}_{2}(\mathrm{g})\) is \(0.2168 \mathrm{Lmol}^{-1}\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). (Note: \(1 \mathrm{MPa}=\) \(\left.1 \times 10^{6} \mathrm{Pa} .\right)\)(a) Use the van der Waals equation to calculate the pressure of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) if the volume is 0.2168 L. What is the \% error in the calculated pressure? The van der Waals constants are \(a=1.382 \mathrm{L}^{2}\) bar \(\mathrm{mol}^{-2}\) and \(b=0.0319 \mathrm{L} \mathrm{mol}^{-1}\) (b) Use the ideal gas equation to calculate the volume of one mole of \(\mathrm{O}_{2}(\mathrm{g})\) at \(280 \mathrm{K}\) and \(10 \mathrm{MPa}\). What is the \% error in the calculated volume?

What is the partial pressure of \(\mathrm{Cl}_{2}(\mathrm{g}),\) in millimeters of mercury, at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm in a gaseous mixture that consists of \(46.5 \% \mathrm{N}_{2}, 12.7 \% \mathrm{Ne},\) and \(40.8 \%\) \(\mathrm{Cl}_{2},\) by mass?

A 132.10 mL glass vessel weighs 56.1035 g when evacuated and \(56.2445 \mathrm{g}\) when filled with the gaseous hydrocarbon acetylene at \(749.3 \mathrm{mmHg}\) and \(20.02^{\circ} \mathrm{C}\) What is the molar mass of acetylene? What conclusion can you draw about its molecular formula?

To establish a pressure of 2.00 atm in a 2.24 L cylinder containing \(1.60 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C},\) (a) add \(1.60 \mathrm{g} \mathrm{O}_{2} ;(\mathrm{b})\) add \(0.60 \mathrm{g} \mathrm{He}(\mathrm{g}) ;(\mathrm{c})\) add \(2.00 \mathrm{g} \mathrm{He}(\mathrm{g})\) (d) release \(0.80 \mathrm{g} \mathrm{O}_{2}(\mathrm{g})\)

A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(23^{\circ} \mathrm{C}\) and a barometric pressure of 751 Torr. The vapor pressure of water at \(23^{\circ} \mathrm{C}\) is \(21 \mathrm{mmHg}\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in the sample collected is (a) \(21 \mathrm{mmHg}_{i}\) (b) 751 Torr; \((\mathrm{c}) 0.96 \mathrm{atm} ;\) (d) \(1.02 \mathrm{atm}\).
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