A 70 kg patient is in need of intravenous iron supplementation. She was ordered to receive a single intravenous 10 mg/kg dose of iron (III) sulfate (ferric sulfate - Fe2(SO4)3 - MW=399.88 g/mol). Iron (III) sulfate was prepared as a stabilized aqueous solution for this patient by dissolving 750 mg of stabilized iron sulfate salt in water to produce 10 mL of solution. How many milliequivalents of iron (Fe 3+) will the patient receive as a single dose?
Assume the salt completely disassociates in solution.
Mass of the patient = 70 kg.
Dosage of Fe(III) required = 10 mg/kg; therefore, the mass of Fe(III) taken by the patient = (70 kg)*(10 mg/1 kg) = 700 mg.
The molar mass of Fe3(SO4)2 is given; calculate the equivalent mass of Fe(III) by dividing the molecular mass by the valence of Fe(III).
Equivalent mass of Fe3(SO4)3 = (399.88 g/mol)/(3) = 133.293 g/eq (we will replace mole by eq since we are dealing with equivalent weight).
Milliequivalents of Fe(III) corresponding to 750 mg = (750 mg)/(133.293 g/eq) = 5.6267 mEq ≈ 5.63 mEq (ans).
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