Physics, asked by Anonymous, 11 months ago

what amount of zn will be required to produce H2 by its action on dillute H2SO4 which will be completely react with oxygen produced by heating 30 gram of KCLo3??.​

Answers

Answered by sanidhya973
7

Answer:

the answer is given in the pictures above.

Attachments:
Answered by Anonymous
1

Explanation:

Explanation:

Start with the only known quantity in this question

m(KClO3)=30lg

n(KClO3)=30lg

⋅1lmol138.55lg=0.217lmol

The balanced equation of the decomposition of potassium perchlorate

2lKClO3(s)Δ−−→2lKCl(s)+3lO2(g)

suggests the stoichiometric relationship

2lmollKClO

3

decomposes to produce

3

l

mol

l

O

2

(

g

)

Hence the amount of oxygen required for the complete combustion of the unknown amount of

H

2

(

g

)

produced would be

n

(

O

2

)

=

0.217

l

mol

l

KClO

3

3

l

mol

l

O

2

2

l

mol

l

KClO

3

=

0.326

l

mol

l

O

2

Oxygen reacts with hydrogen by the equation

1

l

O

2

(

g

)

+

2

l

H

2

(

g

)

*

−−→

2

l

H

2

O

(

g

)

at a

1

:

2

ratio, meaning that the combustion would consume

n

(

H

2

)

=

0.326

l

mol

l

O

2

2

l

H

2

1

l

O

2

=

0.651

lmollH2

of hydrogen. All these

H2

came from the reaction between

Zn

and dilute sulfuric acid

H2SO4

as seen in the following equation

1lZn(s)+H2SO4(aq)→1lH2(g)+ZnSO4(aq)

where for each mole of

H2(g)

produced,

1lmol of Zn

is consumed. That is:

n(Zn)=n(H2)=0.651lmol

Hence the mass of

Znm(Zn)=n(Zn)⋅M(Zn)=42.56lg

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