Chapter 9: Problem 57
Why is the actual yield of a reaction often not equal to the theoretical yield?
Chapter 9: Problem 57
Why is the actual yield of a reaction often not equal to the theoretical yield?
All the tools & learning materials you need for study success - in one app.
Get started for freeConsider the unbalanced chemical equation \(\mathrm{CaC}_{2}+\mathrm{CO} \rightarrow \mathrm{C}+\mathrm{CaCO}_{3}\) When the reaction is complete, \(135.4 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) produced and \(38.5 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) is left over. Assuming the reaction had a \(100 \%\) yield, what were the mass of the two reactants at the beginning of the reaction
The compound \(\mathrm{P}_{4} \mathrm{O}_{10}\) has an empirical formula of \(\mathrm{P}_{2} \mathrm{O}_{5} .\) By what factor will the percent by mass composition differ for each element between these two formulas? Explain your answer.
The thyroid hormone thyroxine has the molecular formula \(\mathrm{C}_{15} \mathrm{H}_{11} \mathrm{NO}_{4} \mathrm{I}_{4} .\) Calculate the mass percent of each element in thyroxine.
The hormone estradiol contains carbon, hydrogen, and oxygen and has a molar mass of approximately \(272 \mathrm{~g} / \mathrm{mol}\). When \(1.15 \mathrm{~g}\) of estradiol is subjected to combustion analysis, the results yield \(79.26 \% \mathrm{C}\) and \(8.88 \% \mathrm{H}\). What is its molecular formula?
Determine the empirical formula of the compound that is \(43.2 \%\) by mass \(\mathrm{K}, 39.1 \%\) by \(\mathrm{mass} \mathrm{Cl}\), and also contains oxygen.
What do you think about this solution?
We value your feedback to improve our textbook solutions.