Question

When 15 gram of sodium
carbonate is addec
to 259 of ethanoic acid the residue in
found to
to weight 30 gram. What man of
carbon dioxide is released during
reaction ?
the​

Answer 1

Answer:

The chemical reaction will be:

NaHCO

3

+CH

3

COOH→CH

3

COONa+H

2

O+CO

2

molarmass:

NaHCO

3

=84

CH

3

COOH=60

CH

3

COONa=82

CO

2

=44

84gNaHCO

3

+60gCH

3

COOH→82gCH

3

COONa+44gCO

2

Moles of NaHCO

3

=

84

6.3

=0.075

Moles of CH

3

COOH=

60

15

=0.25

∴NaHCO

3

is the limited reagent.

Moles of CO

2

formed=0.075

weight of CO

2

=0.075×44=3.3g