Final answer:
Approximately 6.69 grams of anhydrous iron (III) chloride will remain after the complete dehydration of 5 grams of Iron (III) chloride hexahydrate.
Step-by-step explanation:
In this question, we are given 5 grams of Iron (III) chloride hexahydrate and asked to determine how many grams of anhydrous iron (III) chloride will remain after complete dehydration.
To calculate this, we need to find the molar mass of Iron (III) chloride hexahydrate, which is the sum of the molar masses of each element in the compound. We then convert the mass of the hydrated compound to moles using the molar mass, and finally, convert the moles of the hydrated compound to grams of the anhydrous compound using the molar mass of the anhydrous compound.
The molar mass of Iron (III) chloride hexahydrate is calculated as:
- Molar mass of Fe = 55.85 g/mol
- Molar mass of Cl = 35.45 g/mol
- Molar mass of H2O = 18.015 g/mol
Using these values, the molar mass of Iron (III) chloride hexahydrate can be calculated as:
Molar mass = (Molar mass of Fe × 1) + (Molar mass of Cl × 2) + (Molar mass of H2O × 6)
Molar mass = (55.85 g/mol × 1) + (35.45 g/mol × 2) + (18.015 g/mol × 6)
Molar mass = 162.206 g/mol
Now, to convert the mass of the hydrated compound to moles, we use the molar mass:
Moles of hydrated compound = Mass of hydrated compound / Molar mass of hydrated compound
Moles of hydrated compound = 5 g / 162.206 g/mol
Moles of hydrated compound ≈ 0.0308 mol
Finally, to convert the moles of the hydrated compound to grams of the anhydrous compound, we use the molar mass of the anhydrous compound:
Mass of anhydrous compound = Moles of hydrated compound × Molar mass of anhydrous compound
Mass of anhydrous compound = 0.0308 mol × (Molar mass of Fe + 3 × Molar mass of Cl)
Mass of anhydrous compound = 0.0308 mol × (55.85 g/mol + 3 × 35.45 g/mol)
Mass of anhydrous compound ≈ 6.69 g
Therefore, approximately 6.69 grams of anhydrous iron (III) chloride will remain after the complete dehydration of 5 grams of Iron (III) chloride hexahydrate.