Chemistry, asked by ahirwarjeetu5729, 1 year ago

50 g of ZnS are strongly heated in air to effect partial oxidation and the resultant mass weighed 44 g. What is the ratio of ZnO to ZnS in the resulting mixture?

Answers

Answered by pawankumar1252002
4

Answer:

Explanation:

Explanation:

Start by taking a look at the balanced chemical equation that describes this reaction. Zinc sulfide will react with oxygen gas to produce zinc oxide and sulfur dioxide.

2

ZnS

(

s

)

+

3

O

2

(

g

)

2

ZnO

(

s

)

+

2

SO

2

(

g

)

Now, you know that your sample of zinc sulfide undergoes partial oxidation, which means that only a fraction of the mass of zinc sulfide will react to produce zinc oxide.

Notice that zinc sulfide and zinc oxide are present in a

1

:

1

mole ratio because it takes

2

moles of zinc sulfide to produce

2

moles of zinc oxide.

Use the molar masses of the two compounds to convert this mole ratio to a gram ratio.

You will have

1 mole ZnS

1 mole ZnO

=

1

mole ZnS

97.474 g

1

mole ZnS

1

mole ZnO

81.40 g

1

mole ZnO

=

97.474 g ZnS

81.40 g ZnO

So, you know that the reaction produces

81.40 g

of zinc oxide for every

97.474 g

of zinc sulfide that take part in the reaction.

This is equivalent to saying that when

1 g

of zinc sulfide is oxidized, the reaction produces

0.8351 g

of zinc oxide.

In other words, if the reaction consumes

1 g

of zinc sulfide, the total mass of the sample will decrease by

ZnS consumed



1 g

ZnO produced



0.8351 g

=

0.1649 g

In your case, the total mass of the sample decreases by

50 g

44 g

=

6 g

which means that the reaction must have consumed

6

g decrease

1 g ZnS

0.1649

g decrease

=

36.39 g

This implies that the resultant mixture contains

what you start with



50 g ZnS

what is consumed



36.39 g ZnS

=

what remains unconsumed



13.6 g ZnS

Consequently, the resultant mixture also contains

ZnS + ZnO



44 g

ZnS



13.6 g

=

ZnO



30.4 g

Therefore, you can say that the resulting mixture contains zinc oxide and zinc sulfide in a

30.4

:

13.6

gram ratio, which is pretty close to option (D).

The difference between the values was probably caused by the values I used for the molar masses of the two compounds.

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