Answer to Question #259045 in General Chemistry for MCC

Question

A sample of hydrogen gas has a volume of 7.49 L at a pressure of 22.0 atm and a temperature of 32.0°C. What volume of gaseous water is produced by the following reaction at 125.0°C and 0.975 atm, if all the hydrogen gas reacts with iron (III) oxide? 

Answer 1

• The volume of hydrogen gas is 7.49 L.
• The pressure of hydrogen gas is 22.0 atm.
• The temperature of hydrogen gas is 32.0$°c$
• 305.15
• K
• 32.0∘C(32.0+273.15)K=305.15K
• .
• The pressure of gaseous water is 0.975 atm.
• The temperature of gaseous water is 125.0
• ∘
• C
• (
• 125.0
• +
• 273.15
• )
• K
• =
• 398.15
• K
• 125.0∘C(125.0+273.15)K=398.15K
• .

The balanced chemical equation is written as given below:

F

e

2

O

3

(

s

)

+

3

H

2

(

g

)

→

2

F

e

(

s

)

+

3

H

2

O

(

g

)

Fe2O3(s)+3H2(g)→2Fe(s)+3H2O(g)

The number of moles of hydrogen gas is calculated by using the ideal gas equation as given below:

P

V

=

n

R

T

PV=nRT

Where,

• P is the pressure of a gas, i.e., 22.0 atm.
• T is the temperature of a gas, i.e., 305.15 K.
• V is the volume of a gas, i.e.,7.49 L.
• n is the number of moles of a gas.
• R is the universal gas constant (
• R
• =
• 0.0821
• L
• ⋅
• a
• t
• m
• ⋅
• K
• −
• 1
• ⋅
• m
• o
• l
• −
• 1
• )
• (R=0.0821L⋅atm⋅K−1⋅mol−1)
• .

On substituting the values in the ideal gas equation, we get

P

V

=

n

R

T

22.0

a

t

m

×

7.49

L

=

n

×

0.0821

L

⋅

a

t

m

⋅

K

−

1

⋅

m

o

l

−

1

×

305.15

K

PV=nRT22.0atm×7.49L=n×0.0821L⋅atm⋅K−1⋅mol−1×305.15K

On rearranging and simplifying the ideal gas equation, we get

n

H

2

=

22.0

a

t

m

×

7.49

L

0.0821

L

⋅

a

t

m

⋅

K

−

1

⋅

m

o

l

−

1

×

305.15

K

=

164.8

25.05

=

6.58

m

o

l

nH2=22.0atm×7.49L0.0821L⋅atm⋅K−1⋅mol−1×305.15K=164.825.05=6.58mol

According to stoichiometry, 3 mol of hydrogen gas produces 3 mol of water. Therefore, the number of moles of water that can be produced from 6.58 mol of hydrogen gas is calculated as given below:

M

o

l

e

s

o

f

H

2

O

=

6.58

m

o

l

o

f

H

2

×

3

m

o

l

H

2

O

3

m

o

l

H

2

=

6.58

m

o

l

MolesofH2O=6.58molofH2×3molH2O3molH2=6.58mol

Hence the volume of water is calculated by using the formula as given below:

P

V

=

n

R

T

PV=nRT

Where,

• P is the pressure of a gas, i.e., 0.975 atm.
• T is the temperature of a gas, i.e., 398.15 K.
• V is the volume of a gas.
• n is the number of moles of a gas, i.e., 6.58 mol.
• R is the universal gas constant (
• R
• =
• 0.0821
• L
• ⋅
• a
• t
• m
• ⋅
• K
• −
• 1
• ⋅
• m
• o
• l
• −
• 1
• )
• (R=0.0821L⋅atm⋅K−1⋅mol−1)
• .

On substituting the values in the ideal gas equation, we get

P

V

=

n

R

T

0.975

a

t

m

×

V

H

2

O

=

6.58

m

o

l

×

0.0821

L

⋅

a

t

m

⋅

K

−

1

⋅

m

o

l

−

1

×

398.15

K

PV=nRT0.975atm×VH2O=6.58mol×0.0821L⋅atm⋅K−1⋅mol−1×398.15K

On rearranging and simplifying the ideal gas equation, we get

V

H

2

O

=

6.58

m

o

l

×

0.0821

L

⋅

a

t

m

⋅

K

−

1

⋅

m

o

l

−

1

×

398.15

K

0.975

a

t

m

=

215.1

0.975

=

220.6

L

≈

221

L

VH2O=6.58mol×0.0821L⋅atm⋅K−1⋅mol−1×398.15K0.975atm=215.10.975=220.6L≈221L

Thus, the volume of gaseous water is 221 L.

.