Q.3
Find the n factor of underlined compound in following interaction
P4->
H2PO2- + PH3;
MnO2
Answers
Answered by
2
P
4
⟶PH
3
+H
3
PO
3
Oxidation state of P in P
4
=O
Oxidation state of P in PH
3
& H
3
PO
3
=+3
∴ Change of oxidation state of P=3
∴ Change of oxidation state of P
4
=
2
3×4
=
2
12
=6
∴ Equivalent weight of P
4
=
12/2
Molecularwt.
=
12/2
31×4
=20.66
Answered by
0
Answer:
Cu
2
S+KMnO
4
⟶Cu
2+
+SO
2
+Mn
2+
+1 -2 +2 +4
So, n-factor of Cu
2
S=4−(−2)+2×(2−1)=2+6=8.
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