A 0.879g sample of a CaCl2x2H2O/K2C2O4xH2O solid salt mixture is dissolved in 15

Question

A 0.879g sample of a CaCl2x2H2O/K2C2O4xH2O solid salt mixture is dissolved in 150 mL of deionized water, previously adjusted to a pH hat is basic. The precipitate, after having been filtered and air-dried, has a mass of 0/284g. The limiting reactant in the salt mixture was later determined to be CaCl2x2H2O a) How many moles and grams of the excess reactant, K2C2O4xH2O, reacted in the mixture? b) How many grams of the K2C2O4XH2O in the salt mixture remain unreacted (in excess)?

Explanation / Answer

Molecular formula: CaCl2*2H2O(aq) + K2C2O4*H2O(aq) ----> CaC2O4(s) + 2 KCl(aq) + 3 H2O(l) Net-ionic formula: Ca2+(aq) + (C2O4)2-(aq) ---> CaC2O4(s) a) (0.284 g CaC2O4) x [(1 mole CaC2O4)/(128.10 g CaC2O4)] x [(1 mole CaCl2*2H2O)/(1 mole CaC2O4)] = 0.00222 moles CaCl2*2H2O (0.00222 moles CaCl2*2H2O) x [(147.01 g CaCl2*2H2O)/(1 mole)] = 0.326 g CaCl2*2H2O b) This answer means that in the solid mixture there was (0.879 - 0.326)g of K2C2O4H2O or 0.553 g. d) Finally, the mass percent of each reagent can be calculated using the following equation: %mass = (mi/M)x100% where mi is the mass of each sample within the mixture and M is the mass of total solid sample. You should get: 37.1% CaCl2*2H2O and 62.9% K2C2O4H2O

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