16.8g NaHco3 is treated with 12.0 g CH3CooH the residue was found to weight 20 t . determine the mass of co2 escaped in the reaction
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Answered by
1
Answer:
The mass of residue, is CH
3
COONa solution, should be 24g of CH
3
COONa solution.
In this reaction, sodium bicarbonate reacts with acetic acid to produce sodium acetate, carbon dioxide and water
NaHCO
3
+CH
3
COOH⟶CH
3
COONa+CO
2
+H
2
O
According to law of conversation of masses,
total masses of reactants = total masses of products
reactants are 8.4g NaHCO
3
and 20g CH
3
COOH.
total mass of reactants is =28.4g
mass of residue = (28.4g reactants mass) − (4.4g CO
2
)
= 24.0g residue (sodium acetate solution)
mass of residue is thus 24 grams.
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Answer:
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