Question

In a reaction, 15.3 g of sodium carbonate reacted with 16 g of ethanoic acid. The products were 12.2 g of carbon dioxide, 1 g water and 18.1 g of sodium ethanoate. Show that these observations are ...

Answer 1

sodium carbonate(Na2Co3)+ethanoic acid(CH3COOH)===>carbon dioxide(Co2)+water(H2O)+sodium ethanoate(CH3COOH).

15.3g+16g=12.2g+1g+18.1g

31.3g=31.3g

So the reaction always equal to the product

Answer 2

Answer:

Heya !

Here is your 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 !!! : )