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Chapter 6: Problem 113

Chlorine dioxide, \(\mathrm{ClO}_{2}\), is sometimes used as a chlorinating agent for water treatment. It can be prepared from the reaction below: \(\mathrm{Cl}_{2}(\mathrm{g})+4 \mathrm{NaClO}(\mathrm{aq}) \longrightarrow 4 \mathrm{NaCl}(\mathrm{aq})+2 \mathrm{ClO}_{2}(\mathrm{g})\) In an experiment, \(1.0 \mathrm{L} \mathrm{Cl}_{2}(\mathrm{g}),\) measured at \(10.0^{\circ} \mathrm{C}\) and 4.66 atm, is dissolved in 0.750 L of 2.00 M \(\mathrm{NaClO}(\mathrm{aq}) .\) If \(25.9 \mathrm{g}\) of pure \(\mathrm{ClO}_{2}\) is obtained, then what is the percent vield for this experiment?

Short Answer

Expert verified
The percent yield of \(ClO_2(g)\) in this experiment can be calculated using the procedure outlined in steps 1-4.

Step by step solution

01

Calculate number of moles of \(Cl_2(g)\)

Start by calculating the number of moles of \(Cl_2(g)\). The Ideal Gas Law (PV = nRT) can be used here, where P is the pressure (4.66 atm), V is the volume (1.0 L), R is the gas constant (0.0821 atm·L/mol·K) and T is the temperature (283.15 K as we need to convert the degrees celsius to Kelvin by adding 273.15). So the equation becomes: n = PV/RT
02

Calculate moles of \(ClO_2(g)\) expected (theoretical yield)

The stoichiometry of the reaction shows that two moles of \(ClO_2(g)\) are produced per one mole of \(Cl_2(g)\). Hence, the number of moles of \(ClO_2(g)\) that theoretically can be produced is double that of \(Cl_2(g)\). Multiply the moles of \(Cl_2(g)\) obtained from step 1 by 2 to get the theoretical yield in moles of \(ClO_2(g)\).
03

Calculate the actual yield of \(ClO_2(g)\)

For determining the actual yield, convert the mass of \(ClO_2(g)\) obtained (25.9 g) into moles using its molar mass (67.45 g/mol). The formula for this is: moles = mass/molar mass.
04

Calculate percent yield

To calculate the percent yield, divide the actual yield of \(ClO_2(g)\) (from step 3) by the theoretical yield of \(ClO_2(g)\) (from step 2) and then multiply by 100.

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Most popular questions from this chapter

In research that required the careful measurement of gas densities, John Rayleigh, a physicist, found that the density of \(\mathrm{O}_{2}(\mathrm{g})\) had the same value whether the gas was obtained from air or derived from one of its compounds. The situation with \(\mathrm{N}_{2}(\mathrm{g})\) was different, however. The density of \(\mathrm{N}_{2}(\mathrm{g})\) had the same value when the \(\mathrm{N}_{2}(\mathrm{g})\) was derived from any of various compounds, but a different value if the \(\mathrm{N}_{2}(\mathrm{g})\) was extracted from air. In \(1894,\) Rayleigh enlisted the aid of William Ramsay, a chemist, to solve this apparent mystery; in the course of their work they discovered the noble gases. (a) Why do you suppose that the \(\mathrm{N}_{2}(\mathrm{g})\) extracted from liquid air did not have the same density as \(\mathrm{N}_{2}(\mathrm{g})\) obtained from its compounds? (b) Which gas do you suppose had the greater density: \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air or \(\mathrm{N}_{2}(\mathrm{g})\) prepared from nitrogen compounds? Explain. (c) The way in which Ramsay proved that nitrogen gas extracted from air was itself a mixture of gases involved allowing this nitrogen to react with magnesium metal to form magnesium nitride. Explain the significance of this experiment. (d) Calculate the percent difference in the densities at \(0.00^{\circ} \mathrm{C}\) and 1.00 atm of Rayleigh's \(\mathrm{N}_{2}(\mathrm{g})\) extracted from air and \(\mathrm{N}_{2}(\mathrm{g})\) derived from nitrogen compounds. [The volume percentages of the major components of air are \(78.084 \% \mathrm{N}_{2}, 20.946 \% \mathrm{O}_{2}, 0.934 \% \mathrm{Ar},\) and \(0.0379 \% \mathrm{CO}_{2} .\)

A gaseous hydrocarbon that is \(82.7 \%\) C and \(17.3 \%\) H by mass has a density of \(2.35 \mathrm{g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and 752 Torr. What is the molecular formula of this hydrocarbon?

A mixture of \(1.00 \mathrm{g} \mathrm{H}_{2}\) and \(8.60 \mathrm{g} \mathrm{O}_{2}\) is introduced into a 1.500 L flask at \(25^{\circ} \mathrm{C}\). When the mixture is ignited, an explosive reaction occurs in which water is the only product. What is the total gas pressure when the flask is returned to \(25^{\circ} \mathrm{C} ?\) (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is \(23.8 \mathrm{mmHg}\).)

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 volume, in milliliters, occupied by \(89.2 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) at \(37^{\circ} \mathrm{C}\) and \(737 \mathrm{mmHg} ?\)
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