2Al+Zn(NO3)2=2Al(NO3)3+Zn net ionic equation?

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2Al+Zn(NO3)2=2Al(NO3)3+Zn net ionic equation?

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Answer 1 · 2015-11-09T10:15:13

$2 A l^{\circ} (s)$ $2Al^@ (s)$ + $3 Z n^{2+}$ $3Zn^"2+"$ = $2 A l^{3+}$ $2Al^"3+"$ + $3 Z n^{\circ} (s)$ $3Zn^@ (s)$ (net ionic equation)

Explanation:

The equation given above is wrongly balanced, so I have to correct it first.

$A l^{\circ}$ $Al^@$ + $Z n {(N O_{3})}_{2}$ $Zn(NO_3)_2$ = $A l {(N O_{3})}_{3}$ $Al(NO_3)_3$ + $Z n^{\circ}$ $Zn^@$ (unbalanced)

Tallying the atoms based on subscripts,

left side: $A l$ $Al$ = 1; $Z n$ $Zn$ = 1; ${(N O_{3})}_{3}$ $(NO_3)_3$ = 2
right side: $A l$ $Al$ = 1; $Z n$ $Zn$ = 1; ${(N O_{3})}_{3}$ $(NO_3)_3$ = 3

Notice that I am considering the $N O_{3}^{-}$ $NO_3^-$ ion as one "atom" in order to avoid confusing myself.

Balancing the equations we have:

$2 A l^{\circ} (s)$ $color (blue) 2Al^@ (s)$ + $3 Z n {(N O_{3})}_{2}$ $color (red) 3 Zn(NO_3)_2$ = $2 A l {(N O_{3})}_{3}$ $color (green) 2Al(NO_3)_3$ + $3 Z n^{\circ} (s)$ $color (magenta) 3Zn^@ (s)$ (balanced)

left side: $A l$ $Al$ = (1 x $2$ $color (blue) 2$ ) = 2; $Z n$ $Zn$ = (1 x $3$ $color (red) 3$ ) = 3; ${(N O_{3})}_{3}$ $(NO_3)_3$ = (2 x $3$ $color (red) 3$ ) = 6
right side: $A l$ $Al$ = (1x $2$ $color (green) 2$ ) = 2; $Z n$ $Zn$ = (1 x $3$ $color (magenta) 3$ ) = 3; ${(N O_{3})}_{3}$ $(NO_3)_3$ = (3 x $2$ $color (green) 2$ ) = 6

Now that we have the correct balance equation, we must rewrite the equation showing all the ions involved in the reaction.

$2 A l^{\circ} (s)$ $2Al^@ (s)$ + $3 Z n^{2+}$ $3Zn^"2+"$ + $6 N O_{3}^{-}$ $6NO_3^-$ = $2 A l^{3+}$ $2Al^"3+"$ + $6 N O_{3}^{-}$ $6NO_3^-$ + $3 Z n^{\circ} (s)$ $3Zn^@ (s)$

Notice that for the $N O_{3}^{-}$ $NO_3^-$ , I multiply the subscript with the substance's coefficient. Now let us get rid of your spectator ions (ions that are present in both sides of the equation).

$2 A l^{\circ} (s)$ $2Al^@ (s)$ + $3 Z n^{2+}$ $3Zn^"2+"$ + $cancel (6NO_3^-)$ = $2Al^"3+"$ + $cancel (6NO_3^-)$ + $3Zn^@ (s)$

Now let's show the electrons per half-reaction:

$2Al^@ (s)$ = $2Al^"3+"$ + $6e^-$

$3Zn^"2+"$ + $6e^-$ = $3Zn^@ (s)$

Thus, the net ionic equation is

$2Al^@ (s)$ + $3Zn^"2+"$ = $2Al^"3+"$ + $3Zn^@ (s)$

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2Al+Zn(NO3)2=2Al(NO3)3+Zn net ionic equation?

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