Question

$\Huge\rm{{Question}}$
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water.

Class - 9th.
Subject - Science.
Chapter - Atoms And Molecules.
Page no 32.

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Answer 1

Given :-

In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate.

To Find :-

That this reaction is in accordance with the Law of Conservation of mass

Solution :-

Law of Conservation of mass States that " In a chemical equation mass of reactants is always equal to mass of products "

The Given Reaction is ;

Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

At First Calculate mass of Reactants ;

• Mass of Sodium Carbonate = 5.3 g
• Mass of Ethanoic acid = 6 g

Total Mass of Reactants = 5.3 + 6 = 11.3 g ---( i )

Now , Calculate mass of Products ;

• Mass of Sodium Ethanoate = 8.2 g
• Mass of carbon dioxide = 2.2 g
• Mass of Water = 0.9 g

Total Mass of Products = 8.2 + 2.2 + 0.9 = 11.3 g ---( ii )

Now , By ( i ) and ( ii ) . We can say that , the given reaction is in accordance with Law if Conservation of Mass !

Hope it helps :)

Answer 2

Answer:

According to the question,

Na2CO3 + CH3COOH → CO2 + H₂O +

CH3COONa

Mass of reactants = 5.3 + 6 = 11.3 gm Mass of products = 2.2+0.9+8.2 = 11.3 gm

This shows, that during a chemical reaction, the mass of reactant = mass of the product.

Hence, it shows the law of conservation of mass.