Question #7132d

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Question #7132d

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Answer 1 · 2016-01-21T14:51:10

$2 A l B r_{3} + 3 K_{2} S O_{4} \to 6 K B r + A l_{2} {(S O_{4})}_{3}$ $2AlBr_3 +3 K_2SO_4 -> 6KBr + Al_2(SO_4)_3$

Explanation:

Let $a, b, c and d$ $a, b,c and d$ be the number of molecules in the balanced equation as shown below.

$a A l B r_{3} + b K_{2} S O_{4} \to c K B r + d A l_{2} {(S O_{4})}_{3}$ $aAlBr_3 +b K_2SO_4 -> cKBr + dAl_2(SO_4)_3$

Inspection reveals that the on the right hand side of the equation number of radical $(S O_{4})$ $(SO_4)$ is the highest, i.e., $3$ $3$ . Hence take $d$ $d$ as a lowest integer.
Therefore, with $d = 1$ $d=1$ , the equation becomes

$a A l B r_{3} + b K_{2} S O_{4} \to c K B r + A l_{2} {(S O_{4})}_{3}$ $aAlBr_3 +b K_2SO_4 -> cKBr + Al_2(SO_4)_3$
To balance 3 radicals $S O_{4}$ $SO_4$ , on the left hand side $b = 3$ $b=3$ .
With this the equation becomes

$a A l B r_{3} + 3 K_{2} S O_{4} \to c K B r + A l_{2} {(S O_{4})}_{3}$ $aAlBr_3 +3 K_2SO_4 -> cKBr + Al_2(SO_4)_3$
Now balancing $6$ $6$ atoms of $K$ $K$ on the left hand side, $c = 6$ $c=6$
we get

$a A l B r_{3} + 3 K_{2} S O_{4} \to 6 K B r + A l_{2} {(S O_{4})}_{3}$ $aAlBr_3 +3 K_2SO_4 -> 6KBr + Al_2(SO_4)_3$
To find the value of $a$ $a$ , you may either balance $A l or B r$ $Al or Br$ . We obtain $a = 2$ $a=2$ and get the balanced equation as in the Answer.

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Question #7132d

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