How would you balance the chemical equation by adding coefficients as needed?

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How would you balance the chemical equation by adding coefficients as needed?

$N a O H (a q) + N a N O_{2} (a q) + A l (s) + H_{2} O (l) \to N H_{3} (a q) + N a A l O_{2} (a q)$ $NaOH(aq) + NaNO_2(aq) + Al(s) +H_2O(l) -> NH_3(aq) + NaAlO_2(aq)$

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Answer 1 · 2016-09-08T10:19:53

You have described a redox process, for which 1/2 equations are the standard approach.
$2 A l + 4 H_{2} O (l) + N O_{2}^{-} \to 2 A l O_{2}^{-} + N H_{4}^{+} + 2 H_{2} O (l)$ $2Al +4H_2O(l) + NO_2^(-) rarr 2AlO_2^(-) +NH_4 ^++2H_2O(l)$

Explanation:

Aluminum metal is oxidized to aluminate, $A l O_{2}^{-}$ $AlO_2^-$ (in water this is probably better represented by $A l {(O H)}_{4}^{-}$ $Al(OH)_4^-$ but I'll stick with the former):

$A l + 2 H_{2} O (l) \to A l O_{2}^{-} + 3 e^{-} + 4 H^{+}$ $Al +2H_2O(l) rarr AlO_2^(-) +3e^(-) +4H^+$ . $i$ $i$

This is balanced with respect to mass and charge.

Nitrite anion is reduced to ammonia: $N (+ I I I) \to N (- I I I)$ $N(+III) rarr N(-III)$ .

$N O_{2}^{-} + 7 H^{+} + 6 e^{-} \to N H_{3} + 2 H_{2} O (l)$ $NO_2^(-) +7H^(+) + 6e^(-) rarr NH_3 +2H_2O(l)$ . $i i$ $ii$

Again, this is (I think) balanced with respect to mass and charge, and is therefore reasonable.

And thus $2 \times (i) + (i i)$ $2xx(i)+(ii)$ :

$2 A l (s) + 4 H_{2} O (l) + N O_{2}^{-} \to 2 A l O_{2}^{-} + N H_{4}^{+} + 2 H_{2} O (l)$ $2Al(s) +4H_2O(l) + NO_2^(-) rarr 2AlO_2^(-) +NH_4 ^++2H_2O(l)$

I think this is balanced; but don't trust my arithmetic. If you need help with individual steps, ask for it. All I have done is write 2 half equations. Aluminum is oxidized to $A l^{3 +}$ $Al^(3+)$ ; nitrite is reduced to ammonia.

Note that I think basic conditions are required, not acidic condtions. And we could resolve this by adding $1 \times H O^{-}$ $1xxHO^-$ to each side of the equation (i.e. I want to get rid of the ammonium ion, $N H_{4}^{+}$ $NH_4^+$ ).

$2 A l (s) + 4 H_{2} O (l) + N O_{2}^{-} + O H^{-} \to 2 A l O_{2}^{-} + N H_{3} + 3 H_{2} O (l)$ $2Al(s) +4H_2O(l) + NO_2^(-) +OH^(-) rarr 2AlO_2^(-) +NH_3+3H_2O(l)$
Charge is balanced; mass is balanced, so I'm happy.

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How would you balance the chemical equation by adding coefficients as needed?

$N a O H (a q) + N a N O_{2} (a q) + A l (s) + H_{2} O (l) \to N H_{3} (a q) + N a A l O_{2} (a q)$ $NaOH(aq) + NaNO_2(aq) + Al(s) +H_2O(l) -> NH_3(aq) + NaAlO_2(aq)$

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Explanation:

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How would you balance the chemical equation by adding coefficients as needed?

NaOH(aq) + NaNO_2(aq) + Al(s) +H_2O(l) -> NH_3(aq) + NaAlO_2(aq)NaOH(aq)+NaNO2(aq)+Al(s)+H2O(l)→NH3(aq)+NaAlO2(aq)NaOH(aq) + NaNO_2(aq) + Al(s) +H_2O(l) -> NH_3(aq) + NaAlO_2(aq)

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$N a O H (a q) + N a N O_{2} (a q) + A l (s) + H_{2} O (l) \to N H_{3} (a q) + N a A l O_{2} (a q)$ $NaOH(aq) + NaNO_2(aq) + Al(s) +H_2O(l) -> NH_3(aq) + NaAlO_2(aq)$