To find the total mass of the reactants in the reaction
\[ \text{FeCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{Fe(OH)}_3 + 3\text{HCl}, \]
we first need to calculate the mass of each reactant based on the coefficients in the balanced equation and the atomic masses provided in the table.
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Calculate the mass of FeCl₃ (Iron(III) chloride):
- Iron (Fe): 1 atom × 56 amu = 56 amu
- Chlorine (Cl): 3 atoms × 35 amu = 105 amu (since 3 × 35 = 105)
Total mass of FeCl₃ = mass of Fe + mass of Cl \[ = 56 + 105 = 161 , \text{amu} \]
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Calculate the mass of 3H₂O (Water):
- Hydrogen (H): 6 atoms × 1 amu = 6 amu (since 3 × 2 = 6)
- Oxygen (O): 3 atoms × 16 amu = 48 amu
Total mass of H₂O = mass of H + mass of O \[ = 6 + 16 = 18 , \text{amu} \] Therefore, mass of 3H₂O = 3 × mass of H₂O = 3 × 18 = 54 amu.
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Add the masses of the reactants: Total mass of the reactants = mass of FeCl₃ + mass of 3H₂O \[ = 161 + 54 = 215 , \text{amu} \]
Thus, the total mass of the reactants is 215 amu.