Zn+HNO3~~Zn(NO3)2+N2O+HO. .t
Pls balance it using oxidation number method step wise.I got 2HNO3 instead of 10HNO3.
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Zn+HNO3~~Zn(NO3)2+N2O +HO. .t Pls balance it using oxidation ... Pls balance it using oxidation number method step wise.I got 2HNO3 instead of 10HNO3.
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At first write the correct equation:
Zn + HNO 3 -> Zn(NO 3 ) 2 + NO 2 + H 2 O
Here, Zn is in 0 oxidation state which oxidizes to +2 in Zn(NO 3 ) 2 while N exists in +5 O.S in HNO 3 and reduced to +1 in N2O the reaction involved is:
Zn + 2e - -> Zn +2
Similarly, N +5 ->N +1 + 4e -
Since in product there are 2 N then we multiply the second equation by 2.
2 N +5 -> 2 N +1 + 8 e -
Now the no. of electrons is balanced as :
[Zn 0 ->Zn +2 + 2e - ] x4
[2 N +5 + 8 e - ->2 N +1 ]
------------------------------------
4Zn + 2 N +5 -> 4Zn +2 + 2 N +4
Putting the values in molecular form as :
4Zn + 2 HNO3 -> 4Zn(NO3) 2 + N2O
Now, we will balance the N by in right hand side by HNO
4Zn + 10 HNO3 -> 4 Zn(NO3) 2 + N2O
In H and O is balanced by adding water molecules in right hand side.
4Zn + 10 HNO3 -> 4 Zn(NO3) 2 + N2O + 5H2O
Zn + HNO 3 -> Zn(NO 3 ) 2 + NO 2 + H 2 O
Here, Zn is in 0 oxidation state which oxidizes to +2 in Zn(NO 3 ) 2 while N exists in +5 O.S in HNO 3 and reduced to +1 in N2O the reaction involved is:
Zn + 2e - -> Zn +2
Similarly, N +5 ->N +1 + 4e -
Since in product there are 2 N then we multiply the second equation by 2.
2 N +5 -> 2 N +1 + 8 e -
Now the no. of electrons is balanced as :
[Zn 0 ->Zn +2 + 2e - ] x4
[2 N +5 + 8 e - ->2 N +1 ]
------------------------------------
4Zn + 2 N +5 -> 4Zn +2 + 2 N +4
Putting the values in molecular form as :
4Zn + 2 HNO3 -> 4Zn(NO3) 2 + N2O
Now, we will balance the N by in right hand side by HNO
4Zn + 10 HNO3 -> 4 Zn(NO3) 2 + N2O
In H and O is balanced by adding water molecules in right hand side.
4Zn + 10 HNO3 -> 4 Zn(NO3) 2 + N2O + 5H2O
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