Chapter 9: Problem 58
What do we mean by the percent yield of a reaction?
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Chapter 9: Problem 58
What do we mean by the percent yield of a reaction?
These are the key concepts you need to understand to accurately answer the question.
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Can the actual yield ever be greater than the theoretical yield for a chemical reaction?
Aluminum metal burns in chlorine gas to form aluminum chloride. (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting if \(100.0 \mathrm{~g}\) of \(\mathrm{Al}\) and \(5.00\) moles of \(\mathrm{Cl}_{2}\) are used?
Consider the unbalanced chemical equation \(\mathrm{NO}+\mathrm{O}_{2} \rightarrow \mathrm{NO}_{2}\) (a) Balance the equation. (b) Translate the equation into words using the word mole(s) wherever you can. (c) To produce 2 moles of \(\mathrm{NO}_{2}\) by the reaction you just wrote, how many grams of \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) must you combine? (d) What is the theoretical yield in grams of \(\mathrm{NO}_{2}\) ? (e) You carry out the reaction and recover \(22.5 \mathrm{~g}\) of \(\mathrm{NO}_{2}\). What is the percent yield?
Ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with fluorine gas \(\left(\mathrm{F}_{2}\right)\) to form carbon tetrafluoride gas and hydrogen fluoride gas. If \(2.78 \mathrm{~g}\) of ethylene reacted with an excess of fluorine, how many grams of each product could theoretically be produced?
When \(490.0 \mathrm{mg}\) of iron reacts with excess bromine, a mixture of \(\mathrm{FeBr}_{2}\) and \(\mathrm{FeBr}_{3}\) is produced. It is determined that \(35.5 \%\) of the mixture is iron(II) (bromide). What is the total mass of \(\mathrm{FeBr}_{2} / \mathrm{FeBr}_{3}\) mixture produced?
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