Chapter 3: Problem 35
What is the theoretical yield for a reaction, and how does this quantity depend on the limiting reactant?
Chapter 3: Problem 35
What is the theoretical yield for a reaction, and how does this quantity depend on the limiting reactant?
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Get started for freeA given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n},\) where \(n\) is some whole number. Given that $9.03 \times 10^{20}\( molecules of \)\mathrm{XeF}_{n}\( weigh \)0.368 \mathrm{g},$ determine the value for \(n\) in the formula.
The aspirin substitute, acetaminophen $\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}\right),$ is produced by the following three-step synthesis: $$ \mathrm{I} . \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow $$ $$ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \mathrm{II}\quad \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow $$ $$ \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCl}(a q) $$ $$ \mathrm{III.} \quad \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow $$ $$ \mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l) $$ The first two reactions have percent yields of 87\(\%\) and 98\(\%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?
Fungal laccase, a blue protein found in wood-rotting fungi, is 0.390$\% \mathrm{Cu}$ by mass. If a fungal laccase molecule contains four copper atoms, what is the molar mass of fungal laccase?
A compound that contains only carbon, hydrogen, and oxygen is 48.64$\% \mathrm{C}\( and 8.16\)\% \mathrm{H}$ by mass. What is the empirical formula of this substance?
The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide $\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)$ could be produced? $\mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad$ (unbalanced)
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