Question

question 19
1 pts
in the correctly balanced reaction the coefficient for the h₂o is:
__ al(oh)₃ + h₂so₄ → al₂(so₄)₃ + __ h₂o
1
8
6
3
2

Explanation:

Step1: Balance Al atoms

On the right - hand side, there are 2 Al atoms in \(Al_2(SO_4)_3\). So, put a coefficient of 2 in front of \(Al(OH)_3\) on the left - hand side: \(2Al(OH)_3+\_H_2SO_4
ightarrow1Al_2(SO_4)_3+\_H_2O\)

Step2: Balance \(SO_4^{2 -}\) groups

On the right - hand side, there are 3 \(SO_4^{2 -}\) groups in \(Al_2(SO_4)_3\). So, put a coefficient of 3 in front of \(H_2SO_4\) on the left - hand side: \(2Al(OH)_3 + 3H_2SO_4
ightarrow1Al_2(SO_4)_3+\_H_2O\)

Step3: Balance H and O atoms

Count the H and O atoms.

• For H: In \(2Al(OH)_3\), the number of H atoms from \(Al(OH)_3\) is \(2\times3 = 6\), and from \(3H_2SO_4\) is \(3\times2=6\). So, total H atoms on the left - hand side is \(6 + 6=12\).
• For O: In \(2Al(OH)_3\), the number of O atoms (excluding those in \(OH\)) is \(2\times3 = 6\) (from \(Al\) and \(O\) in \(Al(OH)_3\) structure, but when considering \(H_2O\) formation, we can also count by looking at the overall balance).

The right - hand side: In \(Al_2(SO_4)_3\), there are no H atoms contributing to \(H_2O\) from \(Al_2(SO_4)_3\) (H in \(H_2O\) comes from \(H\) in \(Al(OH)_3\) and \(H_2SO_4\)).
The balanced equation is \(2Al(OH)_3+3H_2SO_4 = Al_2(SO_4)_3 + 6H_2O\)

Answer:

C. 6