Available Apr 14 at 12am - May 10 at 11:59pm 27 days Using 18.92 g of PF3 and 18.84 g of XeF4 in the following reaction, determine the
Limiting Reactant.
PF3 + XeF4 ---> PF5 + Xe

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Answer is:     molar solubility of  yttrium fluorid is 4.23·10⁻⁶ m. balanced chemical reaction: yf₃(s)  → y³⁺(aq) + 3f⁻(aq). ksp = 8.62·10⁻²¹. ksp = [y³⁺]  · [f⁻]³.[y³⁺] = [yf₃] = s.[f⁻] = 3[yf₃] = 3s.ksp(yf₃)  = s  · (3s)³.8.62·10⁻²¹ = 27s⁴.s(yf₃)  =  ⁴√(8.62·10⁻²¹  ÷ 27).s(yf₃)  = 0.00000423 m.s(yf₃) = 4.23·10⁻⁶ m.