Question

Question 2 Consider the following reaction: 4 KO2(s) + 2 H2O(1) ® 4 KOH(s) + 3 O2(g) Suppose 83 grams of KO2 are added to 27 grams of H20. What is the limiting reactant and how many grams of KOH is produced (theoretically)? Answer below and show your work on the work sheet. Label your work as #2. HTML

Answer 1

KO2 is limiting reagent and theoretical yield of KOH is 65.46 g .

» Ч КОИ Is) + АКО, s» + 21,Ou 834 30, 27 Weight: molecular man ligmal" 18.0gmel" No. of moles = Weight Molecular mass 1.167 moles 1, 5 тeos But, KO is a No. of moles of KO No. of moles of H₂O Accdiding to the reaction, mokes of ko, is reating will enoles of ho. here, for 1.5 moles of 120 to react , 3 moles of ko, necessary but only 1.167 moles are available Therefore, is a limiting reagent According Zo reaction 4 moles of ko, will give 4 moles of kon Imole of to will 1 - 16 mag of Ko, give 1.167 moles of Koh mass of of moles of KO ихолоn Ради = 65- 46 46 4 Theoretical yuild = 65.46.9 give Imole mple of кои will Н Кои no = - 167 x 56-1)

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