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For the reaction shown, calculate how many grams of oxygen form when each quantity of

Chapter 8, Problem 32P

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QUESTION:

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

\(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

(a) 2.72 g \(\mathrm{KClO}_{3}\)

(b) 0.361g \(\mathrm{KClO}_{3}\)

(c) 83.6 kg \(\mathrm{KClO}_{3}\)

(d) 22.4 mg \(\mathrm{KClO}_{3}\)

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QUESTION:

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts.

\(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

(a) 2.72 g \(\mathrm{KClO}_{3}\)

(b) 0.361g \(\mathrm{KClO}_{3}\)

(c) 83.6 kg \(\mathrm{KClO}_{3}\)

(d) 22.4 mg \(\mathrm{KClO}_{3}\)

ANSWER:

Step 1 of 4

The reaction given

\(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

We have to calculate how many grams of oxygen form when each quantity of reactant completely reacts.

(a) Given: \(2.72\mathrm{\ g\ KClO}_3\)

Find: \(\mathrm{gO}_{2}\)

\(\begin{array}{l}\mathrm{M}\left(\mathrm{KClO}_3\right)=122.55\frac{\mathrm{g}}{\mathrm{mol}}\text{ and }\mathrm{M}\left(\mathrm{O}_2\right)=32\frac{\mathrm{g}}{\mathrm{mol}}\\ \frac{1\ \mathrm{mol\ KClO}_3}{122.55\mathrm{\ g\ KClO}_3}\rightarrow\frac{3\mathrm{\ mol\ O}_2}{2\mathrm{\ mol\ KClO}_3}\rightarrow\frac{32\mathrm{\ gO}_2}{1\ \mathrm{mol\ O}_2}\\ 2.72\ \mathrm{g\ KClO}_3\times\frac{1\mathrm{\ mol\ KClO}_3}{122.55\ \mathrm{g\ KClO}_3}\times\frac{3\ \mathrm{mol\ O}_2}{2\mathrm{\ mol\ KClO}_3}\times\frac{32\mathrm{\ g\ O}_2}{1\mathrm{\ mol\ O}_2}=1.065\mathrm{\ g\ O}_2\end{array}\)

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