Zn + HNO3 ------ Zn(NO3)2 + NO2 + H2O please balance this in method which you have thought . But not in trail method

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Solution

this is a redox reaction and you should start by writing the oxidation and reduction halves and balancing them. Zn goes from 0 to +2 and N goes from +5 to +4.. do you see that? $(N O_{3})$ is -1.. O is -2.
Hence the oxidation state = $+ 5 + (3 \times - 2) = 1$
$N O_{2}$ is neutral. O is -2, $O_{2}$ is -4.
so N is +4) anyway

$Z n^{0} ⟶ Z n^{+ 2} + 2 e^{-}$
$N^{+ 5} + 1 e^{-} ⟶ N^{+ 4}$

now balance the electrons... multiply the bottom equation by 2

$Z n^{0} ⟶ Z n^{+ 2} + 2 e^{-}$
$2 N^{+ 5} + 2 e^{-} ⟶ 2 N^{+ 4}$

add

now fill in counter ions

$Z n + 2 H N O_{3} ⟶ Z n {(N O_{3})}_{2} + 2 N O_{2}$

now fill in the remaining H2O and HNO3 and put a _ in front of them

$Z n + - H N O_{3} + 2 H N O_{3} ⟶ Z n {(N O_{3})}_{2} + 2 N O_{2} + - H_{2} O$

now fill in the _...

$Z n + 2 H N O_{3} + 2 H N O_{3} ⟶ Z n {(N O_{3})}_{2} + 2 N O_{2} + 2 H_{2} O$

let's check..

left side.. 1 zn, 4 H's, 4 N's, 12 O's
right side 1 Zn, 4H's, 4 N's, 12 O's
electrons. 1 Zn loses 2. 2 N's gain $2 \times 1 = 2$
all is well...

now simplify by adding nitric acid together and putting it in the order you wrote above...

$Z n + 2 H N O_{3} ⟶ Z n {(N O_{3})}_{2} + 2 N O_{2} + 2 H_{2} O$

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Similar questions

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A s + H N O_{3} \to H_{3} A s O_{4} + N O_{2} + H_{2} O

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H {N O}_{3}

Z n

Z n + c o n c . H {N O}_{3} ⟶ Z n {({N O}_{3})}_{2} + X + H_{2} O . . . . . . . . . . . . . . . . . . (A)

Z n + d i l . H {N O}_{3} ⟶ Z n {({N O}_{3})}_{2} + Y + H_{2} O . . . . . . . . . . . . . . . . . . . . (B)

A

B

X

Y

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