How would you balance: CaSO4+AlCl3→CaCl2+Al2(SO4)3?

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How would you balance: CaSO4+AlCl3→CaCl2+Al2(SO4)3?

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Answer 1 · 2015-11-07T13:14:10

$3 C a S O_{4}$ $3CaSO_4$ + $2 A l C l_{3}$ $2AlCl_3$ $\to$ $rarr$ $3 C a C l_{2}$ $3CaCl_2$ + $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$

Explanation:

First tally the atoms. Since $S O_{4}^{2-}$ $SO_4^"2-"$ is an ion, I'm going to consider it as one "atom" in order to not confuse myself.

$C a S O_{4}$ $CaSO_4$ + $A l C l_{3}$ $AlCl_3$ $\to$ $rarr$ $C a C l_{2}$ $CaCl_2$ + $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$ (unbalanced)

Based on the subscripts,

left side:
$C a$ $Ca$ = 1
( $S O_{4}$ $SO_4$ ) = 1
$A l$ $Al$ = 1
$C l$ $Cl$ = 3

right side:
$C a$ $Ca$ = 1
( $S O_{4}$ $SO_4$ ) = 3
$A l$ $Al$ = 2
$C l$ $Cl$ = 2

$3 C a S O_{4}$ $color (red) 3CaSO_4$ + $A l C l_{3}$ $AlCl_3$ $\to$ $rarr$ $C a C l_{2}$ $CaCl_2$ + $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$

Let's start balancing the most complicated 'atom', $S O_{4}^{2-}$ $SO_4^"2-"$

left side:
$C a$ $Ca$ = (1 x $3$ $color (red) 3$ ) = 3
( $S O_{4}$ $SO_4$ ) = (1 x $3$ $color (red) 3$ ) = 3
$A l$ $Al$ = 1
$C l$ $Cl$ = 3

right side:
$C a$ $Ca$ = 1
( $S O_{4}$ $SO_4$ ) = 3
$A l$ $Al$ = 2
$C l$ $Cl$ = 2

Since $C a S O_{4}$ $CaSO_4$ is a substance, we need to also multiply the coefficient with its $C a$ $Ca$ atom. Now that there are 3 $C a$ $Ca$ atoms on the left, there must be 3 $C a$ $Ca$ atoms on the right.

$3 C a S O_{4}$ $3CaSO_4$ + $A l C l_{3}$ $AlCl_3$ $\to$ $rarr$ $3 C a C l_{2}$ $color (blue) 3CaCl_2$ + $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$

left side:
$C a$ $Ca$ = (1 x 3) = 3
( $S O_{4}$ $SO_4$ ) = (1 x 3) = 3
$A l$ $Al$ = 1
$C l$ $Cl$ = 3

right side:
$C a$ $Ca$ = (1 x $3$ $color (blue) 3$ ) = 3
( $S O_{4}$ $SO_4$ ) = 3
$A l$ $Al$ = 2
$C l$ $Cl$ = (2 x $3$ $color (blue) 3$ ) = 6

Again notice that since $C a C l_{2}$ $CaCl_2$ is a substance, the coefficient 3 should also be applied to its $C l$ $Cl$ atoms. Since there are 6 atoms of $C l$ $Cl$ on the right, we need to also have the same number of $C l$ $Cl$ atoms on the left.

$3 C a S O_{4}$ $3CaSO_4$ + $2 A l C l_{3}$ $color (green) 2AlCl_3$ $\to$ $rarr$ $3 C a C l_{2}$ $3CaCl_2$ + $A l_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3$

left side:
$C a$ $Ca$ = (1 x 3) = 3
( $S O_{4}$ $SO_4$ ) = (1 x 3) = 3
$A l$ $Al$ = (1 x $2$ $color (green) 2$ ) = 2
$C l$ $Cl$ = (3 x $2$ $color (green) 2$ ) = 6

right side:
$C a$ $Ca$ = (1 x 3) = 3
( $S O_{4}$ $SO_4$ ) = 3
$A l$ $Al$ = 2
$C l$ $Cl$ = (2 x 3) = 6

Again, since $A l C l_{3}$ $AlCl_3$ is a substance, the coefficient should also apply to the bonded $A l$ $Al$ atom.

Now the equation is balanced.

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How would you balance: CaSO4+AlCl3→CaCl2+Al2(SO4)3?

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