If 26.4 g of hydrogen, H2, is produced from 100.0 g of methane, CH4, reacting with excess water as shown, what is the percentage yield? CH4(g) + H2O(g) → CO(g) + 3H2(g)

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:

43.452

step-by-step explanation:

284 times .153 = 43.452

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm