How many moles of C6H12O6 will be produced if 22 grams of CO2 are reacted

60 grams

Explanation:

We have the balanced equation (without state symbols):

6

H

2

O

+

6

C

O

2

→

C

6

H

12

O

6

+

6

O

2

So, we would need six moles of carbon dioxide to fully produce one mole of glucose.

Here, we got  

88

g

of carbon dioxide, and we need to convert it into moles.

Carbon dioxide has a molar mass of  

44

g/mol

. So here, there exist

88

g

44

g

/mol

=

2

mol

Since there are two moles of  

C

O

2

, we can produce  

2

6

⋅

1

=

1

3

moles of glucose  

(

C

6

H

12

O

6

)

.

We need to find the mass of the glucose produced, so we multiply the number of moles of glucose by its molar mass.

Glucose has a molar mass of  

180.156

g/mol

. So here, the mass of glucose produced is

1

3

mol

⋅

180.156

g

mol

≈

60

g

to the nearest whole number.

So, approximately  

60

grams of glucose will be produced.

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