a 0.879-g sample of CaCl2 X 2H2O/K2C2O4 X H2O solid salt mixture is dissolved in

Question

a 0.879-g sample of CaCl2 X 2H2O/K2C2O4 X H2O solid salt mixture is dissolved in 150 mL of deionized, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.284 g. The limiting reactant in the salt mixture was later determined to be CaCl2 X2H2O. a) write the molecular form of the equation for the reaction b) write the net ionic equation for the reaction c) how many moles and grams of CaCl2 X 2H2O reacted in the reaction? d) how many moles and gram of the excess reactant K2C2O4 X H2O reacted in the mixture? e. How many grams of the K2C2O4 X H2O in the salt mixture remain unreacted (in excess)? f. what is the percent by mass of each salt in the mixture?

Explanation / Answer

A 0.879-g sample of a CaCl2.2H2O/K2C2O4H2O solid salt mixture is dissolved in approximately 150mL of deionized water, previously adjusted to a pH that is basic. The precipitate, after havbing been filtered and air-dried, has a mass of 0.284g. The limiting reactant in the salt mixture was later determined to be CaCl2.2H2O. a.write molecuar formula for the equation for the reaction. b.wrtie the net ionic equation. c. how many moles and grams of CaCl2.2H2O reacted in the reaction mixture? d.How many moles and grams of the excess reactant, K2C2O4H2O, reacted in the mixture? e.How many grams of the K2C2O4H2O in the salt mixture remain UNREACTED (in excess)? f.what is the percent by mass of each salt in the mixture? 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) Since the calculations for (c), (d), and (e) are all based on similar conversions I will only solve (c) and leave the rest for you to finish up. (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 This answer means that in the solid mixture there was (0.879 - 0.326)g of K2C2O4H2O or 0.553 g. 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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