Question

balance the following chemical reaction: al₂(so₄)₃(aq) + k₃po₄(aq) → alpo₄(s) + k₂so₄(aq) answer: a al₂(so₄)₃(aq) + 2k₃po₄(aq) → 2alpo₄(s) + 3k₂so₄(aq) b 2al₂(so₄)₃(aq) + 6k₃po₄(aq) → 3alpo₄(s) + 6k₂so₄(aq) c al₂(so₄)₃(aq) + 4k₃po₄(aq) → 4alpo₄(s) + 6k₂so₄(aq) d also₄(aq) + kpo₄(aq) → alpo₄(s) + kso₄(aq)

Explanation:

Step1: Count atoms on both sides

For the un - balanced equation \(Al_2(SO_4)_3(aq)+K_3PO_4(aq)\to AlPO_4(s)+K_2SO_4(aq)\)

• Aluminum (\(Al\)): Left side has \(2\) atoms (\(Al_2(SO_4)_3\)), right side has \(1\) atom (\(AlPO_4\))
• Sulfur (\(S\)): Left side has \(3\) atoms (\(Al_2(SO_4)_3\)), right side has \(1\) atom (\(K_2SO_4\))
• Oxygen (\(O\)): Left side has \(12 + 4=16\) atoms (\(3\times4\) from \(SO_4\) and \(4\) from \(PO_4\) in \(K_3PO_4\)), right side has \(4 + 4 = 8\) atoms (\(4\) from \(AlPO_4\) and \(4\) from \(SO_4\) in \(K_2SO_4\))
• Potassium (\(K\)): Left side has \(3\) atoms (\(K_3PO_4\)), right side has \(2\) atoms (\(K_2SO_4\))
• Phosphorus (\(P\)): Left side has \(1\) atom (\(K_3PO_4\)), right side has \(1\) atom (\(AlPO_4\))

Step2: Balance the equation

Let's start by balancing the sulfate (\(SO_4\)) group.
We have \(3\) \(SO_4\) groups on the left (\(Al_2(SO_4)_3\)). So we need \(3\) \(K_2SO_4\) on the right.
Now for potassium (\(K\)): Since we have \(3\) \(K_2SO_4\), we have \(6\) \(K\) atoms on the right. So we need \(2\) \(K_3PO_4\) on the left to get \(6\) \(K\) atoms (\(2\times3\)).
For aluminum (\(Al\)): We have \(1\) \(Al_2(SO_4)_3\) on the left (\(2\) \(Al\) atoms). So we need \(2\) \(AlPO_4\) on the right.
The balanced equation is \(Al_2(SO_4)_3(aq)+2K_3PO_4(aq)\to 2AlPO_4(s)+3K_2SO_4(aq)\)

Answer:

A. \(Al_2(SO_4)_3(aq)+2K_3PO_4(aq)\to 2AlPO_4(s)+3K_2SO_4(aq)\)