a(NH ) PO b(Al CO ) o 42 3 4 c(NH ) CO dAlPO
The value of a, b, c and d will be
(1) 2, 1, 3, 2 (2) 1, 2, 3, 2 (3) 3, 3, 2, 2 (4) 1, 2, 3, 4
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Answered by
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Answer:
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
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