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Sure, let's start by balancing the given chemical equation and finding the mole ratio of [tex]\( \text{H}_2\text{SO}_4 \)[/tex] to [tex]\( \text{H}_2\text{O} \)[/tex].
The unbalanced chemical reaction is:
[tex]\[ \text{H}_2\text{SO}_4 + \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \][/tex]
### Step-by-Step Balancing the Equation:
#### 1. Balance the aluminum ([tex]\(\text{Al}\)[/tex]) atoms:
On the product side, there are 2 aluminum atoms in [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]. Therefore, we need 2 [tex]\(\text{Al(OH)}_3\)[/tex] on the reactant side:
[tex]\[ \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \][/tex]
#### 2. Balance the sulfate ([tex]\(\text{SO}_4\)[/tex]) groups:
On the product side, there are 3 sulfate ions (from [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]). Therefore, we need 3 [tex]\(\text{H}_2\text{SO}_4\)[/tex] on the reactant side:
[tex]\[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \][/tex]
#### 3. Balance the hydrogen ([tex]\(\text{H}\)[/tex]) and oxygen ([tex]\(\text{O}\)[/tex]) atoms:
Balance the hydrogen and oxygen on both sides. On the reactant side:
- Hydrogen atoms from [tex]\(3 \text{H}_2\text{SO}_4\)[/tex]: [tex]\(3 \times 2 = 6\)[/tex] H atoms
- Hydrogen atoms from [tex]\(2 \text{Al(OH)}_3\)[/tex]: [tex]\(2 \times 3 = 6\)[/tex] H atoms
- Total hydrogen atoms on reactant side: [tex]\(6 + 6 = 12\)[/tex] H atoms
On the product side, the hydrogen atoms must come from water. Therefore, we need:
- [tex]\(6 \text{H}_2\text{O}\)[/tex] to balance the 12 H atoms, as [tex]\(6 \times 2 = 12\)[/tex]
So the correctly balanced equation is:
[tex]\[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O} \][/tex]
### Determining the Mole Ratio:
Now, from the balanced equation:
[tex]\[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O} \][/tex]
We see that 3 moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex] produce 6 moles of [tex]\(\text{H}_2\text{O}\)[/tex]. Therefore, the mole ratio of [tex]\(\text{H}_2\text{SO}_4\)[/tex] to [tex]\(\text{H}_2\text{O}\)[/tex] is:
[tex]\[ \frac{3}{6} \][/tex]
### Simplifying the Quotient:
[tex]\[ \frac{3}{6} = \frac{1}{2} \][/tex]
So, the correct mole ratio of [tex]\(\text{H}_2\text{SO}_4\)[/tex] to [tex]\(\text{H}_2\text{O}\)[/tex] is:
[tex]\[ \boxed{0.5} \][/tex]
The correct answer is [tex]\(0.5\)[/tex].
The unbalanced chemical reaction is:
[tex]\[ \text{H}_2\text{SO}_4 + \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \][/tex]
### Step-by-Step Balancing the Equation:
#### 1. Balance the aluminum ([tex]\(\text{Al}\)[/tex]) atoms:
On the product side, there are 2 aluminum atoms in [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]. Therefore, we need 2 [tex]\(\text{Al(OH)}_3\)[/tex] on the reactant side:
[tex]\[ \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \][/tex]
#### 2. Balance the sulfate ([tex]\(\text{SO}_4\)[/tex]) groups:
On the product side, there are 3 sulfate ions (from [tex]\(\text{Al}_2(\text{SO}_4)_3\)[/tex]). Therefore, we need 3 [tex]\(\text{H}_2\text{SO}_4\)[/tex] on the reactant side:
[tex]\[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \][/tex]
#### 3. Balance the hydrogen ([tex]\(\text{H}\)[/tex]) and oxygen ([tex]\(\text{O}\)[/tex]) atoms:
Balance the hydrogen and oxygen on both sides. On the reactant side:
- Hydrogen atoms from [tex]\(3 \text{H}_2\text{SO}_4\)[/tex]: [tex]\(3 \times 2 = 6\)[/tex] H atoms
- Hydrogen atoms from [tex]\(2 \text{Al(OH)}_3\)[/tex]: [tex]\(2 \times 3 = 6\)[/tex] H atoms
- Total hydrogen atoms on reactant side: [tex]\(6 + 6 = 12\)[/tex] H atoms
On the product side, the hydrogen atoms must come from water. Therefore, we need:
- [tex]\(6 \text{H}_2\text{O}\)[/tex] to balance the 12 H atoms, as [tex]\(6 \times 2 = 12\)[/tex]
So the correctly balanced equation is:
[tex]\[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O} \][/tex]
### Determining the Mole Ratio:
Now, from the balanced equation:
[tex]\[ 3 \text{H}_2\text{SO}_4 + 2 \text{Al(OH)}_3 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6 \text{H}_2\text{O} \][/tex]
We see that 3 moles of [tex]\(\text{H}_2\text{SO}_4\)[/tex] produce 6 moles of [tex]\(\text{H}_2\text{O}\)[/tex]. Therefore, the mole ratio of [tex]\(\text{H}_2\text{SO}_4\)[/tex] to [tex]\(\text{H}_2\text{O}\)[/tex] is:
[tex]\[ \frac{3}{6} \][/tex]
### Simplifying the Quotient:
[tex]\[ \frac{3}{6} = \frac{1}{2} \][/tex]
So, the correct mole ratio of [tex]\(\text{H}_2\text{SO}_4\)[/tex] to [tex]\(\text{H}_2\text{O}\)[/tex] is:
[tex]\[ \boxed{0.5} \][/tex]
The correct answer is [tex]\(0.5\)[/tex].
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