What minimum mass of iron (II)nitrate must be added to 10.0 of a 0.0699 M phosphate solution in order to completely precipitate all of the phosphate as solid iron (II)phosphate? 2PO₄^3-(aq)+ 3Fe(NO₃)₂(aq)→ Fe₃(PO₄)₂(s)+ 6NO₃^-(aq)

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To find the minimum mass of iron (II) nitrate needed, first calculate the moles of phosphate ions: 10.0 L * 0.0699 M = 0.699 moles. According to the balanced equation, 3 moles of Fe(NO3)2 are needed for every 2 moles of PO4^3-. Therefore, 0.699 moles of PO4^3- require 1.0485 moles of Fe(NO3)2. The molar mass of Fe(NO3)2 is approximately 179.86 g/mol, so the mass needed is 1.0485 moles * 179.86 g/mol = 188.57 g, which rounds to 0.188 g.

What Minimum Mass of Iron (II)nitrate Must Be Added to 10.0