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Chapter 22: Problem 32

In the electrolysis of a sample of water \(22.83 \mathrm{mL}\) of \(\mathrm{O}_{2}(\mathrm{g})\) was collected at \(25.0^{\circ} \mathrm{C}\) at an oxygen partial pressure of \(736.7 \mathrm{mmHg} .\) Determine the mass of water that was decomposed.

Short Answer

Expert verified
The calculations should yield the mass of water that was decomposed in the electrolysis process.

Step by step solution

01

Convert volume of Oxygen to moles

The volume of the oxygen collected during electrolysis is given as \(22.83 mL\). This volume needs to be converted to liters because the gas law uses liters. Therefore, divide the volume by 1000 to get it in liters: \( V = 22.83 mL = 22.83 / 1000 = 0.02283 L\). Next, convert the pressure from mmHg to atm using the conversion factor (1 atm/760 mmHg). This gives \(P = 736.7 mmHg = 736.7 / 760 = 0.969 atm\). Lastly, convert the temperature from Celsius to Kelvin (T(K) = T(°C) + 273.15), which gives \(T = 25.0°C = 25.0 + 273.15 = 298.15 K\). The molar gas constant R = 0.0821 L.atm/(mol.K). Apply PV=nRT and solve for n (number of moles): \( n = PV/RT = (0.969 atm * 0.02283 L) / (0.0821 L.atm/mol.K * 298.15 K)\).
02

Use stoichiometry to find moles of water

The balanced equation for the electrolysis of water is \(2H_2O -> 2H_2 + O_2\). This shows that two moles of water produce one mole of oxygen. So, if x is the number of moles of water decomposed, according to the stoichiometric ratio x/2 = n. Therefore, solving for x, x = 2n.
03

Convert moles of water to grams of water

We know that the molar mass of water is approximately 18.015 g/mol. Using this information, convert the number of moles of water to grams of water by multiplying by the molar mass of water to obtain the mass (m) of water decomposed: \( m = x * 18.015 g/mol\).

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Most popular questions from this chapter

Draw Lewis structures for \(\mathrm{O}_{3}\) and \(\mathrm{SO}_{2}\). In what ways are the structures similar? In what ways do they differ?

A 55 L cylinder contains \(A r\) at 145 atm and \(26^{\circ}\) C. What minimum volume of air at STP must have been liquefied and distilled to produce this Ar? Air contains \(0.934 \%\) Ar, by volume.

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What volume of \(\mathrm{H}_{2}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) is required to hydrogenate oleic acid, \(\mathrm{C}_{17} \mathrm{H}_{33} \mathrm{COOH}(1)\) to produce one mole of stearic acid, \(\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COOH}(\mathrm{s}) ?\) Assume reaction (22.52) proceeds with a 95\% yield.

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.
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