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Compounds containing ruthenium(II) and bipyridine.
Chapter 3, Problem 138P(choose chapter or problem)
Compounds containing ruthenium(II) and bipyridine, \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\), have received considerable interest because of their role in systems that convert solar energy to electricity. The compound \({\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right] \mathrm{Cl}_{2}}\) is synthesized by reacting \(\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) with three molar equivalents of \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}(\mathrm{~s})\), along with an excess of triethylamine, \(\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}(\mathrm{l})\), to convert ruthenium(III) to ruthenium(II). The density of triethylamine is 0.73 g/mL, and typically eight molar equivalents are used in the synthesis, (a) Assuming that you start with 6.5 g of \(\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\), how many grams of \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\) and what volume of \(\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}\) should be used in the reaction? (b) Given that the yield of this reaction is 91 percent, how many grams of \({\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right] \mathrm{Cl}_{2}}\) will be obtained?
Questions & Answers
QUESTION:
Compounds containing ruthenium(II) and bipyridine, \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\), have received considerable interest because of their role in systems that convert solar energy to electricity. The compound \({\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right] \mathrm{Cl}_{2}}\) is synthesized by reacting \(\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) with three molar equivalents of \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}(\mathrm{~s})\), along with an excess of triethylamine, \(\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}(\mathrm{l})\), to convert ruthenium(III) to ruthenium(II). The density of triethylamine is 0.73 g/mL, and typically eight molar equivalents are used in the synthesis, (a) Assuming that you start with 6.5 g of \(\mathrm{RuCl}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\), how many grams of \(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\) and what volume of \(\mathrm{N}\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3}\) should be used in the reaction? (b) Given that the yield of this reaction is 91 percent, how many grams of \({\left[\mathrm{Ru}\left(\mathrm{C}_{10} \mathrm{H}_{8} \mathrm{~N}_{2}\right)_{3}\right] \mathrm{Cl}_{2}}\) will be obtained?
ANSWER:
Step 1 of 3
(a)
Here, we are asked to calculate the mass of and volume of used in the reaction.
First, let’s find the number of moles of :
Molar mass of = 261.465 g/mol
Molar mass of = 156.19 g/mol
Molar mass of triethylamine() = 101.19 g/mol.
Number of moles (n) =
=
= 0.0248 mol.
Given, three molar equivalents of are used in the reaction:
= (30.0248 mol)156.19 g/mol
= 11.64 g
Hence, the mass of are used in the reaction is 11.64 g.