How do you balance $N O_{2} + H_{2} O \to H N O_{3} + N O$ $NO_2 + H_2O → HNO_3 + NO$ ?

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Answer 1 · 2017-11-25T13:38:13

$3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O$ $3NO_2+H_2O -> 2HNO_3+NO$

Given:

$N O_{2} + H_{2} O \to H N O_{3} + N O$ $NO_2 + H_2O -> HNO_3 + NO$

We are looking for something like:

$a N O_{2} + b H_{2} O \to c H N O_{3} + d N O$ $aNO_2 + bH_2O -> cHNO_3 + d NO$

for some integers $a, b, c, d$ $a, b, c, d$ .

Equating the amounts of $N$ $N$ , $O$ $O$ and $H$ $H$ , we find:

${ (a = c + d), (2a+b = 3c+d), (2b = c) :}$

Subtracting the first equation from the second to eliminate $d$ , we find:

$a+b = 2c$

Then subtracting twice the third equation, we find:

$a-3b = 0$

So let's put $b=1$ , $a=3$ , $c=2$ and $d=1$ to find:

$3NO_2+H_2O -> 2HNO_3+NO$

How do you balance NO_2 + H_2O → HNO_3 + NONO2+H2O→HNO3+NONO_2 + H_2O → HNO_3 + NO?