To determine the mass of [tex]\( MnO_2 \)[/tex] produced when 445 grams of [tex]\( H_2O \)[/tex] are reacted, let's follow each step carefully using stoichiometry principles.
### Step 1: Determine the moles of [tex]\( H_2O \)[/tex]
To find the moles of [tex]\( H_2O \)[/tex], we need the molar mass of [tex]\( H_2O \)[/tex]. The molar mass of [tex]\( H_2O \)[/tex] is given as 18.01528 g/mol.
[tex]\[ \text{Moles of } H_2O = \frac{\text{Mass of } H_2O}{\text{Molar mass of } H_2O} \][/tex]
[tex]\[ \text{Moles of } H_2O = \frac{445 \, \text{g}}{18.01528 \, \text{g/mol}} = 24.701253602497435 \, \text{mol} \][/tex]
### Step 2: Use the balanced equation to relate moles of [tex]\( H_2O \)[/tex] to moles of [tex]\( MnO_2 \)[/tex]
From the balanced chemical equation:
[tex]\[ H_2O + 2 MnO_4^- + Br^- \rightarrow BrO_3^- + 2 MnO_2 + 2 OH^- \][/tex]
We see that 1 mole of [tex]\( H_2O \)[/tex] produces 2 moles of [tex]\( MnO_2 \)[/tex]. Therefore, the moles of [tex]\( MnO_2 \)[/tex] are calculated as:
[tex]\[ \text{Moles of } MnO_2 = 2 \times \text{Moles of } H_2O \][/tex]
[tex]\[ \text{Moles of } MnO_2 = 2 \times 24.701253602497435 \, \text{mol} = 49.40250720499487 \, \text{mol} \][/tex]
### Step 3: Convert moles of [tex]\( MnO_2 \)[/tex] to grams
The molar mass of [tex]\( MnO_2 \)[/tex] is given as 86.9368 g/mol. To find the mass, use the formula:
[tex]\[ \text{Mass of } MnO_2 = \text{Moles of } MnO_2 \times \text{Molar mass of } MnO_2 \][/tex]
[tex]\[ \text{Mass of } MnO_2 = 49.40250720499487 \, \text{mol} \times 86.9368 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } MnO_2 = 4294.895888379198 \, \text{g} \][/tex]
### Conclusion
The mass of [tex]\( MnO_2 \)[/tex] produced is approximately 4294.90 grams. Comparing this with the given options:
- 5,200 g
- 4,300 g
- 0.430 g
- 8,600 g
The correct answer is:
[tex]\[ 4,300 \, \text{g} \][/tex]
