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Question

Potassium superoxide, KO2KO_2, is employed in a self-contained breathing apparatus used by emergency personnel as a source of oxygen. The reaction is\

4KO2(s)+2H2O(l)4KOH(s)+3O2(g)4KO_2(s) + 2H_2O(l) \longrightarrow 4KOH(s) + 3O_2(g)\

Say a self-contained breathing apparatus is charged with 750 g KO2KO_2 and then is used to produce 195 g of oxygen. Was all of the KO2KO_2 consumed in this reaction? If the KO2KO_2 wasn’t all consumed, how much is left over and what mass of additional O2O_2 could be produced?

Solution

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Answered 2 years ago
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From the problem we have next reaction:

4  KO2  (s)+2  H2O  (l)4  KOH  (s)+3  O2  (g)4 \; \text{KO}_2 \; (\text{s}) + 2 \; \text{H}_2\text{O} \; (\text{l}) \rightarrow 4 \; \text{KOH} \; (\text{s}) + 3 \; \text{O}_2 \; (\text{g})

We have 750 g of KO2\text{KO}_2 and 195 g of oxygen. We need to determine if all of the KO2\text{KO}_2 consumed in this reaction, and If the KO2\text{KO}_2 wasn’t all consumed, how much is left over and what mass of additional O2\text{O}_2 could be produced.

We have:

m(KO2)=750  g\text{m}(\text{KO}_2)=750 \; \text{g} m(O2)=195  g\text{m}(\text{O}_2)=195 \; \text{g}

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