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bruno102
 one year ago
Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) arrow 2Hg (l) + O2 (g). If 3.00 moles of HgO decompose to form 1.25 moles of O2 and 503 g of Hg, what is the percent yield of this reaction?
bruno102
 one year ago
Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) arrow 2Hg (l) + O2 (g). If 3.00 moles of HgO decompose to form 1.25 moles of O2 and 503 g of Hg, what is the percent yield of this reaction?

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bruno102
 one year ago
Best ResponseYou've already chosen the best response.141.6% 62.5% 83.3% 96.9%

pooja195
 one year ago
Best ResponseYou've already chosen the best response.0Do you have any clue on how to do this?

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1I have done others like it but this one is a bit different. I attempted to do this and got about 83.5...

pooja195
 one year ago
Best ResponseYou've already chosen the best response.0Thats what i got....:/

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1Well I am not sure how to go about solving it :/ I guess I will just select 83.3%.

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1@Ciarán95 did you have any ideas?

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1Are you actually typing a reply?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A. 41.1 is the correct answer

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1@Doctor_At_Work Really? how did you get that?

Ciarán95
 one year ago
Best ResponseYou've already chosen the best response.1Firstly, let's define what 'Percentage Yield' is: Percentage Yield = (Actual Yield (in grams)/Theoretical Yield (in grams)) x 100 Our Actual Yield is the amount of a specific product we find to have been produced at the end of the reaction we have carried out, given the amount of our reactant we started with. It seems logical to think that we're being asked for the Percentage Yield of Mercury (Hg), given that we're told that 503g were produced following the decomposition of 3 moles of HgO. However, given this 3 moles of HgO which we start with, we can work out from the stoichiometric/molar ratios that are in the balanced chemical equation how much SHOULD THEORETICALLY BE FORMED. In practice though, product can often be lost though unwanted side reactions, human error, procedure discrepancies and such else, so our Percentage Yield value is a good indicator as to how successful a reaction or synthesis is. Anyway, back to the task at hand! If we look at the numbers (molar coefficients) in front of the HgO, Hg and O2 in the balanced chemical equation, they tell us that two molecules of HgO decompose to produce 2 atoms of Hg and 1 molecule of O2. Or, in other words, given 2 moles of HgO we will produce 2 moles of Hg and 1 mole of O2. So, focusing in on our target product of Mercury, we see that we should always theoretically produce the same number of moles of Hg as the number of moles of HgO we originally had. Therefore, given that the question tells us that 3 moles of HgO are initally there, we would expect, theoretically that 3 moles of Hg liquid forms. To work out the % Yield, we need to convert this into grams for Mercury. If you look on the periodic table, you will see the relative atomic mass of Mercury is 200.59 (element number 80). 1 mole of pure Hg will always contain its relative atomic mass in grams  in other words, 1 mole of Hg will contain 200.59 grams. We define this as the molar mass of Hg, writing it as 200.59 g/mol. So, to convert moles into grams: \[Grams = Moles \times Molar Mass\] So, in our case: Number of grams of Hg = (3 mol) x (200.59 g/mol) = 601.77g This is our Theoretical Yield value for Mercury in this reaction. Now we know both this and the Actual Yield (503g) in grams. So, we can proceed to identify the Percentage Yield: \[Percentage Yield = \frac{ 503g }{ 601.77g } \times 100\] This gives us 83.59%, correct to 4 significant figures, like what @bruno102 got. This doesn't seem to match up exactly with any of the options...I can only assume that we're dealing with the Mercury and not the O2 gas. If that's all the info you're given I would say to select 83.3% as you're answer...the offset may be due to rounding error with the relative atomic mass of Hg. Anyway, hope that helped you out a little bit! :)

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1Thanks that did help! It is how I solved it, the only concern I have is what doctor_at_work mentioned about the answer being 41.1

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1@Ciarán95 you think that @doctor_at_works's answer is wrong?

Ciarán95
 one year ago
Best ResponseYou've already chosen the best response.1Looking back over it again I can't see any way it can be 41.1%...if your method for solving it is the same as mine it certainly seems to be correct. It's for @Doctor_At_Work to support his reasoning really, but if 41.1% was to be correct then the Theoretical Yield would have to be 1223.84g, not 601.77g as we worked out.

Ciarán95
 one year ago
Best ResponseYou've already chosen the best response.1By the way, welcome to OpenStudy @Doctor_At_Work ! :D

bruno102
 one year ago
Best ResponseYou've already chosen the best response.1Yes, I figured it out just as you did. Thank you so much for your help :)
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