When HgO is heated, it decomposes into elemental mercury and molecular oxygen gas. If 66.0 g of Hg is obtained from 82.0 g of the oxide, what is the percent yield of the reaction?
I worked one just like this earlier. Here is the link.
http://www.jiskha.com/display.cgi?id=1418084714
To calculate the percent yield of a reaction, you need to compare the actual yield to the theoretical yield.
First, let's find the theoretical yield of mercury (Hg) using stoichiometry.
The balanced equation for the reaction is:
2 HgO -> 2 Hg + O2
From the balanced equation, we can see that 2 moles of mercury oxide (HgO) produce 2 moles of mercury (Hg). Therefore, the molar ratio between HgO and Hg is 1:1.
1 mole of HgO has a molar mass of 216.59 g/mol.
So, 82.0 g of HgO is equivalent to (82.0 g) / (216.59 g/mol) = 0.3786 moles of HgO.
Since the molar ratio is 1:1, the theoretical yield of Hg is also 0.3786 moles.
The molar mass of mercury (Hg) is 200.59 g/mol.
Therefore, the theoretical yield of Hg is (0.3786 moles) * (200.59 g/mol) = 75.96 g.
Now, we can calculate the percent yield.
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Given that the actual yield of Hg is 66.0 g, we have:
Percent Yield = (66.0 g / 75.96 g) * 100 = 86.6%
So, the percent yield of the reaction is 86.6%.