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Chapter 22: Problem 52

A 25.0 L sample of a natural gas, measured at \(25^{\circ} \mathrm{C}\). and 740.0 Torr, is bubbled through \(\mathrm{Pb}^{2+}(\mathrm{aq}),\) yielding \(0.535 \mathrm{g}\) of \(\mathrm{PbS}(\mathrm{s}) .\) What mass of sulfur can be recovered per cubic meter of this natural gas?

Short Answer

Expert verified
For every cubic meter of the natural gas, 2.86 g of sulfur can be recovered.

Step by step solution

01

Convert volume to cubic meters

First, we need to convert the volume of the gas from liters to cubic meters (m³). Remember there are 1000L in 1m³. So, 25.0 L equals \(0.025 \, m³\).
02

Determine the number of moles of PbS(s)

By using the molar mass of PbS(s), which is 239.3g/mol, calculate the number of moles in 0.535 gram of PbS(s). By dividing 0.535g by the molar mass of PbS(s), we get 0.00223 mol of PbS(s).
03

Use stoichiometry to find moles of sulfur

From the chemical formula of PbS(s), we know that for each molecule of PbS(s), there is one atom of sulfur, S(s). Therefore, the number of moles of S(s) will be equal to the number of moles of PbS(s), which is 0.00223 mol.
04

Calculate the mass of sulfur

The molar mass of sulfur (S) is approximately 32.06g/mol. To calculate the mass of sulfur, multiply the number of moles (0.00223 mol) by its molar mass. Therefore, the mass of sulfur in the 25.0 L sample is approximately \(0.0715 g\).
05

Find the mass of sulfur per cubic meter of gas

We know the mass of sulfur in 0.025 m³ of natural gas, and we are asked to find the mass per m³. By dividing the obtained mass of sulfur (0.0715g) by the volume of natural gas in cubic meter (0.025 m³), the sulfur content per cubic meter of this natural gas is \(2.86 \, g/m³\).

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Most popular questions from this chapter

Use VSEPR theory to predict the probable geometric structures of the molecules (a) \(\mathrm{O}_{2} \mathrm{XeF}_{2} ;\) (b) \(\mathrm{O}_{3} \mathrm{XeF}_{2}\) (c) OXeF \(_{4}\).

Use information from this chapter and previous chapters to write plausible chemical equations to represent the following: (a) the reaction of silver metal with \(\mathrm{HNO}_{3}(\mathrm{aq})\) (b) the complete combustion of the rocket fuel, unsymmetrical dimethylhydrazine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\) (c) the preparation of sodium triphosphate by heating a mixture of sodium dihydrogen phosphate and sodium hydrogen phosphate.

Use the following electrode potential diagram for basic solutions to classify each of the statements below as true or false. Assume standard conditions. $$\begin{aligned}\mathrm{SO}_{4}^{2-} \stackrel{-0.936 \mathrm{V}}{\longrightarrow} \mathrm{SO}_{3}^{2-} & \stackrel{-0.576 \mathrm{V}}{\longrightarrow} \\\& \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \stackrel{-0.74 \mathrm{V}}{\longrightarrow} \mathrm{S} \stackrel{-0.476 \mathrm{V}}{\longrightarrow} \mathrm{S}^{2-}\end{aligned}$$ (a) Sulfate \(\left(\mathrm{SO}_{4}^{2-}\right)\) is a stronger oxidant than thiosulfate \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2}\right)\) in basic solution. (b) \(S^{2-}\) can be used as a reducing agent in basic solutions. (c) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is stable with respect to disproportionation to \(\mathrm{SO}_{3}^{2-}\) and \(\mathrm{S}\) in basic solution.

Polonium is the only element known to crystallize in the simple cubic form. In this structure, the interatomic distance between a Po atom and each of its six nearest neighbors is \(335 \mathrm{pm}\). Use this description of the crystal structure to estimate the density of polonium.

Make a general prediction about which of the halogen elements, \(\mathrm{F}_{2}, \mathrm{Cl}_{2}, \mathrm{Br}_{2},\) or \(\mathrm{I}_{2},\) displaces other halogens from a solution of halide ions. Which of the halogens is able to displace \(\mathrm{O}_{2}(\mathrm{g})\) from water? Which is able to displace \(\mathrm{H}_{2}(\mathrm{g})\) from water?
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