. Potassium superoxide, KO2 is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l) - 4KOH(s) + 3O2(g) If a reaction vessel contains 0.15 mol KO2 and 0.10 mol H2O, what is the limiting reactant? How many moles of oxygen can be produced? 

Dear Student,


4KO2(s) + 2H2O(l)  4KOH(s) +3O2(g)4 mol of KO2 reacts with 2 mol of H2OSo, 0.158 mol of KO2 will react with 24×0.158 =0.079 mol of H2OBut we have 0.10 mol of H2O, so KO2 is the limiting reagent.Now, 4 mol of KO2 gives 3 mol of O2So, 0.158 mol of KO2 will give 34×0.158= 0.1185 mol of O2. 

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In the equation, 4 mole of KO2 reacts with 2 mole of H2O. =1 mole of KO2 will react with 2÷4 mole of H2O. =0.15 mole of KO2 will react with 2÷4×0.15 mole of H2O. =0.15 mole of KO2 will react with 0.07 mole of H2O. But we have 0.10 mole of H2O. Thus, KO2 is the limiting reagent. Again according to the equation, 4mole of KO2 produces 3mole of O2. =0.15 mole of KO2 will produce 3÷4×0.15 mole of O2. =0.15 mole of KO2 will produce 0.11 mole of O2. Ans
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