To calculate the total mass of the reactants in the reaction:
\[ \text{FeCl}_3 + 3\text{H}_2\text{O} \]
We need to find the mass of all the elements in the reactants individually.
-
FeCl₃:
- Mass of Fe = 56 amu
- Mass of Cl = 35 amu (and there are 3 Cl atoms in FeCl₃)
\[ \text{Total mass of FeCl}_3 = 56 + (3 \times 35) = 56 + 105 = 161 \text{ amu} \]
-
3H₂O:
- Mass of H = 1 amu (and there are 2 H atoms in each water molecule, so for 3 water molecules we have 6 H atoms)
- Mass of O = 16 amu (and there is 1 O atom in each water molecule, so for 3 water molecules we have 3 O atoms)
\[ \text{Total mass of } 3\text{H}_2\text{O} = (3 \times 2 \times 1) + (3 \times 16) = 6 + 48 = 54 \text{ amu} \]
Now, add the total mass of the reactants together:
\[ \text{Total mass of reactants} = \text{Total mass of FeCl}_3 + \text{Total mass of } 3\text{H}_2\text{O} = 161 + 54 = 215 \text{ amu} \]
Therefore, the total mass of the reactants is 215 amu.