In a reaction 5.5 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.4 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. Hint (sodium carbonate + ethanoic acid → sodium ethanoate +carbon dioxide + water)
Answers
Answer:
In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water
sodium carbonate + ethanoic acid †’ sodium ethanoate + carbon dioxide + water
Mass of sodium carbonate=5.3g (given)
Mass of ethanoic acid=6g(given)
Mass of carbon dioxide=2.2g(given)
Mass of water=0.9g(given)
Mass of sodium ethanoate=8.2g(given)
Now total mass before reaction=(5.3g+6g)=11.3g
Now total mass after reaction=(8.2g+2.2g+0.9g)=11.3g
Total mass before reaction=total mass after reaction
Therefore the given observation is in agreement with law of conservation of mass.
Explanation:
Answer:
In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water
sodium carbonate + ethanoic acid †’ sodium ethanoate + carbon dioxide + water
Mass of sodium carbonate=5.3g (given)
Mass of ethanoic acid=6g(given)
Mass of carbon dioxide=2.2g(given)
Mass of water=0.9g(given)
Mass of sodium ethanoate=8.2g(given)
Now total mass before reaction=(5.3g+6g)=11.3g
Now total mass after reaction=(8.2g+2.2g+0.9g)=11.3g
Total mass before reaction=total mass after reaction
Therefore the given observation is in agreement with law of conservation of mass.
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