Question

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%

Answer 1

#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.