If the percentage yield of a reaction is 85%

% 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

Explanation:

no answer,

Explanation: but if the other dude see this can you re word Wht u said it’s hard to make sense of it

b) mutualism

hope this : )