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For the reaction shown, calculate how many grams of oxygen form when each quantity of
Chapter 8, Problem 32P(choose chapter or problem)
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 \(\mathrm{KClO}_{3}\)
Questions & Answers
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: \(\mathrm{gO}_{2}\)
\(\begin{array}{l}\mathrm{M}\left(\mathrm{KClO}_3\right)=122.55\frac{\mathrm{g}}{\mathrm{mol}}\text{ and }\mathrm{M}\left(\mathrm{O}_2\right)=32\frac{\mathrm{g}}{\mathrm{mol}}\\ \frac{1\ \mathrm{mol\ KClO}_3}{122.55\mathrm{\ g\ KClO}_3}\rightarrow\frac{3\mathrm{\ mol\ O}_2}{2\mathrm{\ mol\ KClO}_3}\rightarrow\frac{32\mathrm{\ gO}_2}{1\ \mathrm{mol\ O}_2}\\ 2.72\ \mathrm{g\ KClO}_3\times\frac{1\mathrm{\ mol\ KClO}_3}{122.55\ \mathrm{g\ KClO}_3}\times\frac{3\ \mathrm{mol\ O}_2}{2\mathrm{\ mol\ KClO}_3}\times\frac{32\mathrm{\ g\ O}_2}{1\mathrm{\ mol\ O}_2}=1.065\mathrm{\ g\ O}_2\end{array}\)