Week 2 Workshop Question

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Marisa Gaitan 2D
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Week 2 Workshop Question

Postby Marisa Gaitan 2D » Mon Oct 26, 2020 8:54 am

Hi, I believe this is a question from a workshop a couple weeks ago. I wasn't able to make it so I couldn't see the solutions. Would anyone be able to walk me through your thought process. I was able to get an answer, but I'm not sure if I did it correctly.

Potassium superoxide reacts with water to produce oxygen and potassium hydroxide in the following reaction: KO2​ (s) + H​2O (l) → O​2 (g) + KOH (s)
a. 8.54 g of KO​2 (s) reacts with 27.7 g of liquid water, and the percent yield of KOH (s) is found to be 59.7%. How much KOH (s) was produced, in grams?
b. If the actual yield of KOH (s) from part (a) is dissolved in 8.67 L of water, what is the concentration of the KOH (aq) solution? Assume 100% of the KOH dissolves.

Josh Chou 3K
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Re: Week 2 Workshop Question

Postby Josh Chou 3K » Mon Oct 26, 2020 11:55 am

Hi! So the first thing I did was write a balanced equation: 4KO2(s) + 2H2O(l) --> 3O2(g) + 4KOH(aq). Then, I converted the given masses into moles, giving me 0.12 moles of KO2 and 1.53 moles of H2O. Because 4 moles of KO2 are used for every 2 moles of H2O and there is significantly less KO2, KO2 is the limiting reactant. Because the ratio of KO2 to KOH is 1:1, 0.12 moles of KO2 will have theoretical yield of 0.12 moles KOH, which is 6.73g after converting moles to grams. Only 59.7% of the theoretical yield was actually produced, meaning that 4.02g KOH was the experimental yield. Then, to find the concentration of the solution in B, I converted 4.02g KOH to moles of KOH and then divided that result by 8.67L, giving me a solution with a concentration of 0.008M. I hope this helps! And it would be great if someone could double check my work

Chem_Mod
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Re: Week 2 Workshop Question

Postby Chem_Mod » Mon Oct 26, 2020 1:13 pm

Josh Chou 1I wrote:Hi! So the first thing I did was write a balanced equation: 4KO2(s) + 2H2O(l) --> 3O2(g) + 4KOH(aq). Then, I converted the given masses into moles, giving me 0.12 moles of KO2 and 1.53 moles of H2O. Because 4 moles of KO2 are used for every 2 moles of H2O and there is significantly less KO2, KO2 is the limiting reactant. Because the ratio of KO2 to KOH is 1:1, 0.12 moles of KO2 will have theoretical yield of 0.12 moles KOH, which is 6.73g after converting moles to grams. Only 59.7% of the theoretical yield was actually produced, meaning that 4.02g KOH was the experimental yield. Then, to find the concentration of the solution in B, I converted 4.02g KOH to moles of KOH and then divided that result by 8.67L, giving me a solution with a concentration of 0.008M. I hope this helps! And it would be great if someone could double check my work


Your work is correct Josh!


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