Let's work through this problem step-by-step to find out how many moles of [tex]\( \text{H}_2\text{O} \)[/tex] are produced when 3.45 moles of [tex]\( \text{KMnO}_4 \)[/tex] are used.
First, let's analyze the equation given:
[tex]\[ 3.45 \, \text{mol} \, \text{KMnO}_4 \times \frac{8 \, \text{mol} \, \text{H}_2\text{O}}{2 \, \text{mol} \, \text{KMnO}_4} \][/tex]
This equation indicates that for every 2 moles of [tex]\( \text{KMnO}_4 \)[/tex], 8 moles of [tex]\( \text{H}_2\text{O} \)[/tex] are produced. To proceed, follow these steps:
### Step 1: Determine the mole ratio between [tex]\( \text{KMnO}_4 \)[/tex] and [tex]\( \text{H}_2\text{O} \)[/tex]
The stoichiometric ratio provided is:
[tex]\[ \frac{8 \, \text{mol} \, \text{H}_2\text{O}}{2 \, \text{mol} \, \text{KMnO}_4} \][/tex]
This ratio can be simplified:
[tex]\[ \frac{8}{2} = 4 \][/tex]
### Step 2: Use the mole ratio to find the number of moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced
Next, multiply the initial moles of [tex]\( \text{KMnO}_4 \)[/tex] by the simplified ratio:
[tex]\[ 3.45 \, \text{mol} \, \text{KMnO}_4 \times 4 = 13.8 \, \text{mol} \, \text{H}_2\text{O} \][/tex]
### Conclusion
Thus, using 3.45 moles of [tex]\( \text{KMnO}_4 \)[/tex] will produce 13.8 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
