How do you balance $N H_{3} (g) + O_{2} (g) \to N_{2} (g) + H_{2} O (g)$ $NH_3(g) + O_2(g) -> N_2(g) + H_2O(g)$ ?

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How do you balance $N H_{3} (g) + O_{2} (g) \to N_{2} (g) + H_{2} O (g)$ $NH_3(g) + O_2(g) -> N_2(g) + H_2O(g)$ ?

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Answer 1 · 2016-01-07T18:28:22

$4 N H_{3} (g) + 3 O_{2} (g) \to 2 N_{2} (g) + 6 H_{2} O (g)$ $4NH_3(g) + 3O_2(g) → 2N_2(g) + 6 H_2O(g)$

Explanation:

This can be easily balanced through trial and error method or through algebraic method. Always consider to take inventories of the atoms from each element of the reactants and the products; that is, before and after it is balanced.

Balancing through trial and error method:

Reactants:

$N=cancel(1)4$
$H=cancel(3)12$
$O=cancel(2)6$

Products:

$N=cancel(2)4$
$H=cancel(2)12$
$O=cancel(1)6$

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How do you balance $N H_{3} (g) + O_{2} (g) \to N_{2} (g) + H_{2} O (g)$ $NH_3(g) + O_2(g) -> N_2(g) + H_2O(g)$ ?

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How do you balance NH_3(g) + O_2(g) -> N_2(g) + H_2O(g)NH3(g)+O2(g)→N2(g)+H2O(g)NH_3(g) + O_2(g) -> N_2(g) + H_2O(g)?

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How do you balance $N H_{3} (g) + O_{2} (g) \to N_{2} (g) + H_{2} O (g)$ $NH_3(g) + O_2(g) -> N_2(g) + H_2O(g)$ ?