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To determine how many moles of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] are produced from the complete reaction of 449 grams of [tex]\( \text{FeBr}_3 \)[/tex], we can follow these steps:
1. Calculate the molar mass of [tex]\( \text{FeBr}_3 \)[/tex]:
- The molar mass of iron (Fe) is 55.845 g/mol.
- The molar mass of bromine (Br) is 79.904 g/mol.
- Therefore, the molar mass of [tex]\( \text{FeBr}_3 \)[/tex] is calculated as follows:
[tex]\[ \text{Molar mass of } \text{FeBr}_3 = 55.845 + 3 \times 79.904 = 295.557 \, \text{g/mol} \][/tex]
2. Calculate the molar mass of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex]:
- The molar mass of sulfur (S) is 32.06 g/mol.
- Therefore, the molar mass of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] is:
[tex]\[ \text{Molar mass of } \text{Fe}_2\text{S}_3 = 2 \times 55.845 + 3 \times 32.06 = 207.87 \, \text{g/mol} \][/tex]
3. Calculate the number of moles of [tex]\( \text{FeBr}_3 \)[/tex] given the mass:
- Given mass of [tex]\( \text{FeBr}_3 \)[/tex] is 449 grams.
- Using the molar mass of [tex]\( \text{FeBr}_3 \)[/tex]:
[tex]\[ \text{Moles of } \text{FeBr}_3 = \frac{449 \, \text{g}}{295.557 \, \text{g/mol}} = 1.5191655078377435 \, \text{mol} \][/tex]
4. Determine the stoichiometric relationship between [tex]\( \text{FeBr}_3 \)[/tex] and [tex]\( \text{Fe}_2\text{S}_3 \)[/tex]:
- According to the balanced chemical equation:
[tex]\[ 2 \, \text{FeBr}_3 + 3 \, \text{Na}_2\text{S} \rightarrow \text{Fe}_2\text{S}_3 + 6 \, \text{NaBr} \][/tex]
- 2 moles of [tex]\( \text{FeBr}_3 \)[/tex] produce 1 mole of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex].
5. Calculate the number of moles of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] produced:
- For every 2 moles of [tex]\( \text{FeBr}_3 \)[/tex], 1 mole of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] is produced.
- Therefore, the moles of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] produced is:
[tex]\[ \text{Moles of } \text{Fe}_2\text{S}_3 = \frac{1.5191655078377435}{2} = 0.7595827539188718 \, \text{mol} \][/tex]
Thus, the complete reaction of 449 grams of [tex]\( \text{FeBr}_3 \)[/tex] would produce 0.7595827539188718 moles of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex].
1. Calculate the molar mass of [tex]\( \text{FeBr}_3 \)[/tex]:
- The molar mass of iron (Fe) is 55.845 g/mol.
- The molar mass of bromine (Br) is 79.904 g/mol.
- Therefore, the molar mass of [tex]\( \text{FeBr}_3 \)[/tex] is calculated as follows:
[tex]\[ \text{Molar mass of } \text{FeBr}_3 = 55.845 + 3 \times 79.904 = 295.557 \, \text{g/mol} \][/tex]
2. Calculate the molar mass of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex]:
- The molar mass of sulfur (S) is 32.06 g/mol.
- Therefore, the molar mass of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] is:
[tex]\[ \text{Molar mass of } \text{Fe}_2\text{S}_3 = 2 \times 55.845 + 3 \times 32.06 = 207.87 \, \text{g/mol} \][/tex]
3. Calculate the number of moles of [tex]\( \text{FeBr}_3 \)[/tex] given the mass:
- Given mass of [tex]\( \text{FeBr}_3 \)[/tex] is 449 grams.
- Using the molar mass of [tex]\( \text{FeBr}_3 \)[/tex]:
[tex]\[ \text{Moles of } \text{FeBr}_3 = \frac{449 \, \text{g}}{295.557 \, \text{g/mol}} = 1.5191655078377435 \, \text{mol} \][/tex]
4. Determine the stoichiometric relationship between [tex]\( \text{FeBr}_3 \)[/tex] and [tex]\( \text{Fe}_2\text{S}_3 \)[/tex]:
- According to the balanced chemical equation:
[tex]\[ 2 \, \text{FeBr}_3 + 3 \, \text{Na}_2\text{S} \rightarrow \text{Fe}_2\text{S}_3 + 6 \, \text{NaBr} \][/tex]
- 2 moles of [tex]\( \text{FeBr}_3 \)[/tex] produce 1 mole of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex].
5. Calculate the number of moles of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] produced:
- For every 2 moles of [tex]\( \text{FeBr}_3 \)[/tex], 1 mole of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] is produced.
- Therefore, the moles of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex] produced is:
[tex]\[ \text{Moles of } \text{Fe}_2\text{S}_3 = \frac{1.5191655078377435}{2} = 0.7595827539188718 \, \text{mol} \][/tex]
Thus, the complete reaction of 449 grams of [tex]\( \text{FeBr}_3 \)[/tex] would produce 0.7595827539188718 moles of [tex]\( \text{Fe}_2\text{S}_3 \)[/tex].
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