2)
For a theoretical yield of 19 g and actual
yield of 11 g, calculate the percent yield for a
chemical reaction.
Answer in units of %.
11 g / 19 g times 100 = 57.89%
in your 2 sigfig problem, you might be expected to round off to 58%
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3)
In the reaction below, 15 g of H2S with excess
O2 produced 6 g of sulfur.
2H2S + 1O2 --> 1 S + 2H2O.
find moles, using molar mass:
15 g H2S @ 34.08 g/mol = 0.440 moles H2S
0.440 moles H2S produces 0.440 moles S
0.440 mol S @ 32.066 g/mol = 14.11 grams of Sulfur is expected
% yield =
6 g S / 14.11 g theo times 100 = 42.51 % yield
in a 2 sig fig problem this rounds to 43%
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What is the percent 5)
The reaction of 8.3 grams of chlorine with
excess fluorine produced 8.6 grams of ClF3.
What percent yield of ClF3 was obtained?
using molar masses:
8.3 g Cl @ (92.45 g/mol ClF3) / (35.45 g/mol Cl) = 21.65 grams of ClF3 is expected
% yield:
8.3 / 21.65 times 100 = 38.35%
your 2 sig fig problem rounds that to 38%
#1 is ok.
#2 is not. Theoretical yield is ok but actual yield is 8.6; therefore,
(8.6/21.65)*100 = ? and round to 2 places.
To calculate percent yield in a chemical reaction, you need to know the theoretical yield (the amount of product that would be obtained if the reaction went to completion) and the actual yield (the amount of product actually obtained).
To calculate percent yield, you can use the following formula:
Percent Yield = (Actual Yield / Theoretical Yield) x 100
Let's go through the process of calculating percent yield for each of the given examples:
Example 2:
The theoretical yield is given as 19 g, and the actual yield is given as 11 g.
Percent Yield = (11 g / 19 g) x 100 = 57.89%
Since the answer should have 2 significant figures, the result would be rounded to 58%.
Example 3:
The reaction is given as:
2H2S + 1O2 -> 1S + 2H2O
First, calculate the moles of H2S:
Given mass of H2S = 15 g
Molar mass of H2S = 34.08 g/mol
Moles of H2S = (15 g / 34.08 g/mol) = 0.440 mol
Since the stoichiometry of the reaction is 2:1 for H2S and sulfur, the moles of sulfur produced will also be 0.440 mol.
Next, calculate the expected mass of sulfur:
Molar mass of S = 32.066 g/mol
Expected mass of sulfur = (0.440 mol x 32.066 g/mol) = 14.11 g
The actual yield is given as 6 g.
Percent Yield = (6 g / 14.11 g) x 100 = 42.51%
Rounded to 2 significant figures, the result would be 43%.
Example 5:
The reaction is given as:
8.3 g Cl + excess F2 -> 8.6 g ClF3
First, calculate the expected mass of ClF3:
Given mass of Cl = 8.3 g
Molar mass of ClF3 = 92.45 g/mol
Molar mass of Cl = 35.45 g/mol
Expected mass of ClF3 = (8.3 g x 92.45 g/mol) / 35.45 g/mol = 21.65 g
The actual yield is given as 8.6 g.
Percent Yield = (8.6 g / 21.65 g) x 100 = 39.74%
Rounded to 2 significant figures, the result would be 38%.