To calculate the total mass of the reactants in the equation:
\[ \text{FeCl}_3 + 3\text{H}_2\text{O} \]
we need to find the masses of all the reactants.
-
Mass of \( \text{FeCl}_3 \):
- 1 Iron (Fe) = 56 amu
- 3 Chlorine (Cl) = 3 * 35 amu = 105 amu
- Total mass of \( \text{FeCl}_3 = 56 + 105 = 161 \) amu
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Mass of \( 3\text{H}_2\text{O} \):
- For 1 \( \text{H}_2\text{O} \):
- 2 Hydrogen (H) = 2 * 1 amu = 2 amu
- 1 Oxygen (O) = 16 amu
- Total mass of \( \text{H}_2\text{O} = 2 + 16 = 18 \) amu
- Mass of \( 3\text{H}_2\text{O} = 3 * 18 = 54 \) amu
- For 1 \( \text{H}_2\text{O} \):
-
Total mass of the reactants: \[ \text{Total mass} = \text{Mass of } \text{FeCl}_3 + \text{Mass of } 3\text{H}_2\text{O} = 161 \text{ amu} + 54 \text{ amu} = 215 \text{ amu} \]
Thus, the total mass of the reactants is 215 amu.