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Author Question: According to the following balanced reaction, how many moles of KOH will be formed from 5.44 moles ... (Read 168 times)

ahriuashd

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on: Nov 18, 2019

Question 1

According to the following balanced reaction, how many moles of KO are required to exactly react with 5.44 moles of H2O?

4KO(s) + 2H2O(l) → 4KOH(s) + O2(g)

◦ 4.87 moles KO
◦ 8.33 moles KO
◦ 10.9 moles KO
◦ 16.7 moles KO
◦ 2.72 moles KO

Question 2

According to the following balanced reaction, how many moles of KOH will be formed from 5.44 moles of H2O? Assume an excess of KO.

4KO(s) + 2H2O(l) → 4KOH(s) + O2(g)

◦ 2.72 moles KOH
◦ 16.7 moles KOH
◦ 8.33 moles KOH
◦ 4.87 moles KOH
◦ 10.9 moles KOH
Chemistry: A Molecular Approach (Canadian) ¦ Tro, Fridgen, Shaw ¦ 2nd Edition
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Marked as best answer by ahriuashd on Nov 18, 2019

at

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Reply #1 on: Nov 18, 2019
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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ahriuashd

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Reply 2 on: Nov 18, 2019
Excellent

cici

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Reply 3 on: Yesterday
Gracias!

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