posted by Kim

A 453 mL solution has a chloride concentration of 0.0253 M. This solution was prepared by adding solid FeCl3 to water. How many grams of FeCl3 are required?

  1. DrBob222

    mols Cl^- = M x L = approx 0.011 mols.
    0.011 mol Cl^- x (1 mol FeCl3/3 mols Cl^-) = approx 0.004
    Then grams = mols FeCl3 x molar mass FeCl3 = ?

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