Which of the following is the correctly balanced chemical equation for the reaction: Al2(SO4)3 (aq) + K3PO4 (aq) → AlPO4 (s) + K2SO4 (aq)?

• A. Al2(SO4)3 (aq) + 2 K3PO4 (aq) → 2 AlPO4 (s) + 3 K2SO4 (aq)
• B. Al2(SO4)3 (aq) + K3PO4 (aq) → 2 AlPO4 (s) + 3 K2SO4 (aq)
• C. Al2(SO4)3 (aq) + 2 K3PO4 (aq) → AlPO4 (s) + 3 K2SO4 (aq)
• D. Al2(SO4)3 (aq) + 3 K3PO4 (aq) → 2 AlPO4 (s) + 3 K2SO4 (aq)

Answer: (A)

Verified step by step guidance

1. Identify the elements and polyatomic ions involved in the reaction: aluminum (Al), sulfate (SO4), potassium (K), phosphate (PO4), and the compounds Al2(SO4)3, K3PO4, AlPO4, and K2SO4.
2. Write down the unbalanced chemical equation: \(\mathrm{Al_2(SO_4)_3 (aq) + K_3PO_4 (aq) \rightarrow AlPO_4 (s) + K_2SO_4 (aq)}\).
3. Balance the aluminum (Al) atoms first. Since there are 2 Al atoms in \(\mathrm{Al_2(SO_4)_3}\), place a coefficient of 2 in front of \(\mathrm{AlPO_4}\) to balance Al: \(\mathrm{Al_2(SO_4)_3 + K_3PO_4 \rightarrow 2 AlPO_4 + K_2SO_4}\).
4. Balance the phosphate (PO4) groups next. There are 2 phosphate groups on the product side (from \$2 \times \mathrm{AlPO_4}\(), so place a coefficient of 2 in front of \)\mathrm{K_3PO_4}\( on the reactant side: \)\mathrm{Al_2(SO_4)_3 + 2 K_3PO_4 \rightarrow 2 AlPO_4 + K_2SO_4}$.
5. Finally, balance the sulfate (SO4) and potassium (K) ions. There are 3 sulfate groups in \(\mathrm{Al_2(SO_4)_3}\), so place a coefficient of 3 in front of \(\mathrm{K_2SO_4}\) on the product side to balance sulfate and potassium atoms: \(\mathrm{Al_2(SO_4)_3 + 2 K_3PO_4 \rightarrow 2 AlPO_4 + 3 K_2SO_4}\).