When balanced using smallest, whole-number coefficients, the correct coefficients for the following equation are:
_ $A l C l_{3} +$ _ $H_{2} S O_{4} \to$ _ $A l_{2} (S O_{4})_{3} +$ _ $H C l$

2, 1, 1, 2

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2, 3, 1, 6

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1, 3, 1, 3

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2, 3, 1, 3

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Solution

The correct option is D $2, 3, 1, 6$
$A l C l_{3} + H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + H C l$
The products has $2 m o l e s$ of Aluminium, so the coefficient of $A l C l_{3}$ will be 2.
$2 A l C l_{3} + H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + H C l$
The products has $3 m o l e s$ of sulphate, so the coefficient of $H_{2} S O_{4}$ will be 3.
$2 A l C l_{3} + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + H C l$
The reactant has $6 m o l e s$ of chlorine, so the coefficient of $H C l$ will be 6.
$2 A l C l_{3} + 3 H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + 6 H C l$

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