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Chapter 9: Problem 4
How is percentage yield calculated?
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Why is it necessary to use mole ratios in solving stoichiometry problems?
Nitrogen monoxide, NO, reacts with oxygen to form nitrogen dioxide. Then the nitrogen dioxide reacts with oxygen to form nitrogen monoxide and ozone. Write the balanced equations. What is the theoretical yield in grams of ozone from 4.55 g of nitrogen monoxide with excess \(\mathrm{O}_{2} ?\) (Hint: First calcu- late the theoretical yield for NO \(_{2}\) , then use that value to calculate the yield for ozone.)
Iron and CO are made by heating 4.56 \(\mathrm{kg}\) of iron ore, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) and carbon. The yield of iron is 88\(\% .\) How many kilograms of iron are made?
The percentage yield of nitric acid is 95\(\% .\) If 9.88 \(\mathrm{kg}\) of nitrogen dioxide react, what mass of nitric acid is isolated? $$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$$
Use the balanced equation below to write mole ratios for the situations that follow. $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$$ \begin{equation}\begin{array}{l}{\text { a. calculating mol } \mathrm{H}_{2} \mathrm{O} \text { given mol } \mathrm{H}_{2}} \\ {\text { b. calculating mol } \mathrm{O}_{2} \text { given mol } \mathrm{H}_{2} \mathrm{O}} \\ {\text { c. calculating mol } \mathrm{H}_{2} \text { given mol } \mathrm{O}_{2}}\end{array}\end{equation}
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