Question

Sodium carbonate reacts with ethanoic acid to form sodium ethanoate, carbon dioxide and water. In an experiment, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid to form 8.2 g of sodium ethanoate, 2.2 g of carbon dioxide and 0.9 g of water. Show that this data verifies the law of conservation of mass.

Answer 1

Answer:

>> Mass of sodium carbonate = 5.3 g (given)

Mass of ethanoic acid = 6 g (given)

Mass of sodium ethanoate = 8.2 g (given)

Mass of Carbon dioxide = 2.2 g (given)

Mass of water = 0.9 g (given)

Now, total mass before the reaction = (5.3 + 6) g

= 11 . 3 g

And, total mass after the reaction = (8.2 + 2.2 + 0.9 ) g

= 11.3 g

total mass before the reaction = total mass after the reaction.

Hence, the given observations are in agreement with the law of conservation of mass.

Hope it helps !!! : )

Answer 2

Explanation:

Step 1:

Given:

Mass of sodium carbonate = 5.3 g

Mass of ethanoic acid = 6 g

Mass of sodium ethanoate = 8.2 g

Mass of carbon dioxide = 2.2 g

Mass of water = 0.9 g

Step 2:

Total mass (Reactants) = Mass of sodium carbonate + Mass of ethanoic acid

                                       = 5.3 g + 6 g  

                                       = 11.3 g

Step 3:

Total mass (Products) = Mass of sodium ethanoate + Mass of carbon dioxide + mass of water

                                     = 8.2 g + 2.2 g + 0.9 g  

                                     = 11.3 g

In this chemical reaction, Reactants of mass = Product of mass  

Thus it verifies law of conservation of mass.

   

You can get all the answers of Chapter 6 in the link below:     

https://brainly.in/question/15094407

   

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