Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) yields 2Hg (l) + O2 (g). If 4.00
moles of HgO decompose to form 1.50 moles of O2 and 603 g of Hg, what is the percent yield of this reaction?
75%
posted just incase anyone else is looking for the answer^
To find the percent yield of a reaction, we need to compare the actual yield (the amount of product obtained in the experiment) to the theoretical yield (the maximum amount of product that can be obtained based on stoichiometry).
In this case, we are given that 4.00 moles of HgO decompose to form 1.50 moles of O2 and 603 g of Hg. We can use this information to calculate the theoretical yield of mercury (Hg).
According to the balanced equation, the molar ratio between HgO and Hg is 1:1. Therefore, if 4.00 moles of HgO decomposes, we expect to obtain 4.00 moles of Hg.
Now, we can calculate the molar mass of Hg:
atomic weight of Hg = 200.59 g/mol
Next, we can calculate the theoretical yield, which is the mass of Hg that corresponds to the calculated moles:
theoretical yield of Hg = (4.00 moles) x (200.59 g/mol) = 802.36 g
The actual yield of Hg obtained in the experiment is given as 603 g.
Finally, we can calculate the percent yield using the formula:
percent yield = (actual yield / theoretical yield) x 100
percent yield = (603 g / 802.36 g) x 100 = 75.1%
Therefore, the percent yield of this reaction is 75.1%.