![fifthward92](/avatars/23615.jpg)
fifthward92
30.12.2021 •
Business
Which crypto category will have the most growth potential
Solved
Show answers
More tips
- O Other How to Choose the Best Answer to Your Question on The Grand Question ?...
- L Leisure and Entertainment History of International Women s Day: When Did the Celebration of March 8th Begin?...
- S Style and Beauty Intimate Haircut: The Reasons, Popularity, and Risks...
- A Art and Culture When Will Eurovision 2011 Take Place?...
- S Style and Beauty How to Choose the Perfect Hair Straightener?...
- F Family and Home Why Having Pets at Home is Good for Your Health...
- H Health and Medicine How to perform artificial respiration?...
- H Health and Medicine 10 Tips for Avoiding Vitamin Deficiency...
- F Food and Cooking How to Properly Cook Buckwheat?...
- F Food and Cooking How Many Grams Are In a Tablespoon?...
Ответ:
Percent yield represents the ratio between what is experimentally obtained and what is theoretically calculated, multiplied by 100%.
% yield
=
actual yield
theoretical yield
⋅
100
%
So, let's say you want to do an experiment in the lab. You want to measure how much water is produced when 12.0 g of glucose (
C
6
H
12
O
6
) is burned with enough oxygen.
C
6
H
12
O
6
+
6
O
2
→
6
C
O
2
+
6
H
2
O
Since you have a
1
:
6
mole ratio between glucose and water, you can determine how much water you would get by
12.0
g glucose
⋅
1 mole glucose
180.0 g
⋅
6 moles of water
1 mole glucose
⋅
18.0 g
1 mole water
=
7.20
g
This represents your theoretical yield. If the percent yield is 100%, the actual yield will be equal to the theoretical yield. However, after you do the experiment you discover that only 6.50 g of water were produced.
Since less than what was calculated was actually produced, it means that the reaction's percent yield must be smaller than 100%. This is confirmed by
% yield
=
6.50 g
7.20 g
⋅
100
%
=
90.3
%
You can backtrack from here and find out how much glucose reacted
65.0 g of water
⋅
1 mole
18.0 g
⋅
1 mole glucose
6 moles water
⋅
180.0 g
1 mole glucose
=
10.8
g
So not all the glucose reacted, which means that oxygen was not sufficient for the reaction - it acted as a limiting reagent.
Explanation: