If 3.55 mol of HgO decompose to form 1.54 mol of O2. What is the percent yield of this reaction? 2 HgO(s) = 2 Hg(l) + O2(g)
If the theorical yield of a reaction is 23.0g of product and the percent yield is 83%, how many grams were actually produced?
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3.55 mole of `HgO` decomposes to form 1.54 mole of `O_2` .
The chemical equation representing the decomposition of `HgO` into `Hg` and `O_2` is:
`2 HgO(s) -> 2Hg(l) + O_2(g)`
Each mole of `HgO` gives 1 mole of `O_2` . When 3.55 moles of `HgO` decompose, the reaction yields 1.54 mole of `O_2` while it should theoretically give 1.775 moles of `O_2` .
The percent yield of the reaction is `(1.54/1.775)*100` = 86.76%
The theoretical yield of a reaction is 23 g and the percent yield is 83%. In this this case the grams actually produced are equal to `23*(83/100)` = 19.09 g
2 HgO (s) -> 2 Hg (l) + O2 (g)
First, determine the actual mol of O2 that you would get when 3.55 mol of HgO decomposes:
1 mol O2 / 2 mol HgO * 3.55 mol HgO
= 1.78 mol O2
Determine what percent 1.54 mol is of 1.78 mol:
1.54 / 1.78
= 86.5 %
83 % of 23.0 g of a product is:
23.0 g * 0.83
= 19.09 g
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