Calculate the mass of carbon dioxide released in the reaction when 6.3 gm of sodium bicarbonate is added to 15.0 gm of ethanoic acid solution and 13.0 gm of residue left behind
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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
Explanation:
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