How do you balance the chemical equation: $F e + H_{2} O = F e_{3} O_{4} + H_{2}$ $Fe+H_2O = Fe_3O_4+H_2$ ?

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How do you balance the chemical equation: $F e + H_{2} O = F e_{3} O_{4} + H_{2}$ $Fe+H_2O = Fe_3O_4+H_2$ ?

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Answer 1 · 2016-04-28T11:12:40

$3 F e + 4 H_{2} O = F e_{3} O_{4} + 4 H_{2}$ $3Fe+4H_2O = Fe_3O_4+4H_2$

Explanation:

The $F e_{3}$ $Fe_3$ in $F e_{3} O_{4}$ $Fe_3O_4$ on the right implies that there must be some multiple of $3$ $3$ $F e$ $Fe$ on the left side.

Similarly the $O_{4}$ $O_4$ on the right side implies that there must be some multiple of $4$ $4$ $H_{2} O$ $H_2O$ on the left side.

$(3 m) F e + (4 n) H_{2} O = (p) F e_{3} O_{4} + (q) H_{2}$ $(3m)Fe+(4n)H_2O = (p)Fe_3O_4+(q)H_2$

Since
$XXX (3 m) F e \to (m) F e_{3} O_{4}$ $color(white)("XXX")(3m)Fe rarr (m)Fe_3O_4$ (there's no place else for the $F e$ $Fe$ to go)
and
$XXX (4 n) H_{2} O \to (n) F e_{3} O_{4}$ $color(white)("XXX")(4n)H_2O rarr (n)Fe_3O_4$ (there's no place else for the $O$ $O$ to go)
$\Rightarrow m = n$ $rArr m=n$

Further
$XXX (4 n) H_{2} O \to (q) H_{2}$ $color(white)("XXX")(4n)H_2O rarr (q)H_2$
$\Rightarrow q = 4 n$ $rArr q=4n$

Therefore we have
$XXX (3 n) F e + (4 n) H_{2} O = (n) F e_{3} O_{4} + (4 n) H_{2}$ $color(white)("XXX")(3n)Fe+(4n)H_2O = (n)Fe_3O_4+(4n)H_2$

Using the simplest version: $n = 1$ $n=1$ gives
$XXX 3 F e + 4 H_{2} O = F e_{3} O_{4} + 4 H_{2}$ $color(white)("XXX")3Fe+4H_2O = Fe_3O_4+4H_2$

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How do you balance the chemical equation: $F e + H_{2} O = F e_{3} O_{4} + H_{2}$ $Fe+H_2O = Fe_3O_4+H_2$ ?

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Explanation:

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How do you balance the chemical equation: Fe+H_2O = Fe_3O_4+H_2Fe+H2O=Fe3O4+H2Fe+H_2O = Fe_3O_4+H_2?

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How do you balance the chemical equation: $F e + H_{2} O = F e_{3} O_{4} + H_{2}$ $Fe+H_2O = Fe_3O_4+H_2$ ?