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.​

Answer 1

Answer:

Explanation:

Solution

Let us start from the reactant side

In the reactant side we have sodium carbonate and ethanoic acid

Mass of sodium carbonate=5.3g (given)

Mass of ethanoic acid=6g(given)

Now total mass before reaction that is mass of reactants=(5.3g+6g)=11.3g

Now let us check product side

Mass of carbon dioxide=2.2g(given)

Mass of water=0.9g(given)

Mass of sodium ethanoate=8.2g(given)

Now total mass after reaction that is mass of products =(8.2g+2.2g+0.9g)=11.3g

Conclusion

Total mass before reaction=total mass after reaction

Therefore the given observation is in agreement with the law of conservation of mass.

Answer 2

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

We need to show that the law of conservation of mass is agreed in the above reaction.

Solution

Let us start from the reactant side

In the reactant side we have sodium carbonate and ethanoic acid

Mass of sodium carbonate=5.3g (given)

Mass of ethanoic acid=6g(given)

Now total mass before reaction that is mass of reactants=(5.3g+6g)=11.3g

Now let us check product side

Mass of carbon dioxide=2.2g(given)

Mass of water=0.9g(given)

Mass of sodium ethanoate=8.2g(given)

Now total mass after reaction that is mass of products =(8.2g+2.2g+0.9g)=11.3g

Conclusion

Total mass before reaction=total mass after reaction

Therefore the given observation is in agreement with the law of conservation of mass.