To find the total mass of the reactants in the equation \( \text{FeCl}_3 + 3\text{H}_2\text{O} \), we need to calculate the total mass of each reactant:
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For \(\text{FeCl}_3\):
- 1 atom of Fe = 56 amu
- 3 atoms of Cl = 3 × 35 amu = 105 amu
- Total mass of \(\text{FeCl}_3\) = 56 amu + 105 amu = 161 amu
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For \(3\text{H}_2\text{O}\):
- 2 atoms of H in one water molecule = 2 × 1 amu = 2 amu
- 1 atom of O in one water molecule = 16 amu
- Total mass of one water molecule = 2 amu + 16 amu = 18 amu
- For 3 water molecules: \(3 × 18 \text{ amu} = 54 \text{ amu}\)
Now, we can add the total masses of both reactants:
- Total mass of reactants = Mass of \(\text{FeCl}_3\) + Mass of \(3\text{H}_2\text{O}\) = 161 amu + 54 amu = 215 amu
Therefore, the total mass of the reactants is 215 amu.