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

nanotechnology is a powerful tool in biomaterials research as it creates numerous possibilities for the engineering of functional molecules. nano-bio refers to projected ability to incorporate biomimicry into various materials at nanoscale.  

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.