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According to the law of conservation of mass, how much zinc was present in the zinc carbonate?

40 g


88 g


104 g


256 g

According to the law of conservation of mass, how much zinc was present in the zinc-example-1
User Liesel
by
8.0k points

1 Answer

3 votes

Answer : The correct option is, 104 grams.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,


Ca+ZnCO_3\rightarrow CaCO_3+Zn

According to the law of conservation of mass,

Total mass of reactant side = Total mass of product side

Total mass of
Ca+ZnCO_3 = Total mass of
CaCO_3+Zn

Mass of
Ca + Mass of
ZnCO_3 = Mass of
CaCO_3 + Mass of
Zn

As we are given :

Mass of
Ca = 64 g

Mass of
ZnCO_3 = 192 g

Mass of
CaCO_3 = 152 g

So,

Mass of
Ca + Mass of
ZnCO_3 = Mass of
CaCO_3 + Mass of
Zn

64 g + 192 g = 152 g + Mass of
Zn

Mass of
Zn = (64 + 192) - 152

Mass of
Zn = 104 g

Therefore, the mass of zinc present in the zinc carbonate was 104 grams.

User Oldbeamer
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