Answer to Question #280785 in General Chemistry for mayii

Question

When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are  produced. Determine the enthalpy change per mole of zinc reacting for the  reaction: 

 Zn(s) +2HCl (aq) → ZnCl2 (aq) +H2 (g)

Answer 1

• Molatiy of HCl is 0.750 M.
• Volume of HCl is 60.0 mL.
• Mass of Zinc is 1.34 g.
• Molar mass of Zinc is 65.38 g.
• Heat released is 3.14 kJ.

Calculations

Number of moles of HCl can be easily calculate as given below

n

u

m

b

e

r

o

f

m

o

l

e

s

o

f

H

C

l

=

M

o

l

a

r

i

t

y

o

f

H

C

l

×

v

o

l

u

m

e

o

f

H

C

l

n

u

m

b

e

r

o

f

m

o

l

e

s

o

f

H

C

l

=

0.750

m

o

l

L

−

1

×

60

m

L

n

u

m

b

e

r

o

f

m

o

l

e

s

o

f

H

C

l

=

45

m

i

l

lim

o

l

numberofmolesofHCl=MolarityofHCl×volumeofHClnumberofmolesofHCl=0.750molL−1×60mLnumberofmolesofHCl=45millimol

Number of moles of zinc can be easily calculated as shown below.

n

u

m

b

e

r

o

f

m

o

l

e

s

o

f

z

i

n

c

=

m

a

s

s

o

f

z

i

n

c

m

o

l

a

r

m

a

s

s

o

f

z

i

n

c

n

u

m

b

e

r

o

f

m

o

l

e

s

o

f

z

i

n

c

=

1.34

g

65.38

g

n

u

m

b

e

r

o

f

m

o

l

e

s

o

f

z

i

n

c

=

0.0205

m

o

l

n

u

m

b

e

r

o

f

m

o

l

e

s

o

f

z

i

n

c

=

20.5

m

i

l

l

i

m

o

l

numberofmolesofzinc=massofzincmolarmassofzincnumberofmolesofzinc=1.34g65.38gnumberofmolesofzinc=0.0205molnumberofmolesofzinc=20.5millimol

Z

n

+

2

H

C

l

→

Z

n

C

l

2

+

H

2

Zn+2HCl→ZnCl2+H2

In above reaction one mole of zinc reacts with two moles of HCl

It is clearly visible that sufficient amount of HCl is present to consume zinc completely.

We can easily calculate heat released for 1 mole of zinc as shown below.

0.0205

m

o

l

z

i

n

c

r

e

l

e

a

s

e

s

→

3.14

k

J

e

n

e

r

g

y

1

m

o

l

z

i

n

c

w

i

l

l

r

e

l

e

a

s

e

→

3.14

k

J

0.0205

e

n

e

r

g

y

1

m

o

l

z

i

n

c

w

i

l

l

r

e

l

e

a

s

e

→

153.17

k

J

e

n

e

r

g

y

0.0205molzincreleases→3.14kJenergy1molzincwillrelease→3.14kJ0.0205energy1molzincwillrelease→153.17kJenergy

Hence,-153.17 kJ is the value of enthalpy changed per mole of zinc in given reaction.