RRB ALP CBT I 25 Nov 2024 Shift 2 Paper

Concept:A balanced chemical equation has equal numbers of each type of atom on both sides, satisfying the law of conservation of mass.Explanation:Consider the unbalanced reaction:$$\mathrm{Al}_2(\mathrm{SO}_4)_3 + \mathrm{BaCl}_2 \rightarrow \mathrm{AlCl}_3 + \mathrm{BaSO}_4$$Aluminium sulfate contains 2 aluminium atoms and 3 sulfate groups.To balance aluminium, place 2 before $\mathrm{AlCl}_3$ on the product side.To balance sulfate, place 3 before $\mathrm{BaSO}_4$ on the product side.Now place 3 before $\mathrm{BaCl}_2$ on the reactant side to balance barium and chlorine atoms.Count atoms on both sides:Left side: $\mathrm{Al}=2$, $\mathrm{S}=3$, $\mathrm{O}=12$, $\mathrm{Ba}=3$, $\mathrm{Cl}=6$.Right side: $\mathrm{Al}=2$, $\mathrm{S}=3$, $\mathrm{O}=12$, $\mathrm{Ba}=3$, $\mathrm{Cl}=6$.All atoms are equal on both sides, so the balanced equation is:$$\mathrm{Al}_2(\mathrm{SO}_4)_3 + 3\mathrm{BaCl}_2 \rightarrow 2\mathrm{AlCl}_3 + 3\mathrm{BaSO}_4$$This matches option D.Answer:$$\mathrm{Al}_2(\mathrm{SO}_4)_3 + 3\mathrm{BaCl}_2 \rightarrow 2\mathrm{AlCl}_3 + 3\mathrm{BaSO}_4$$Option D is correct.