A 0.879-g sample of a CaCl2x2H2O/K2C2O4xH2O solid salt mixture is dissolved in 1
ID: 599776 • Letter: A
Question
A 0.879-g sample of a CaCl2x2H2O/K2C2O4xH2O solid salt mixture is dissolved in 150 mL of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of .284 g. The limiting reactant in the salt mixture was later determined to be CaCl2x2H2O.a. How many moles and grams of CaCl2*2H2O reacted in the reaction mixture?
b. How many moles and grams of the excess reactant, K2C2O4*H2O, reacted in the mixture?
c. How many grams of the K2CaO4*H2O in the salt mixture remain unreacted?
d. What is the percent by mass of each salt in the mixture?
Explanation / Answer
as CaCl2*2H2O was limiting reactant in the salt mixture, then it should had a mass of 0.284 g so, mass of K2C2O4*H2O should be (0.879 - 0.284= 0.595gm) Moleculer weight of CaCl2*2H2O is 146 gm/mole Moleculer weight of is K2C2O4*H2O 184 gm/mole mole= weight of compound / molecular weight % w/w = [weight of compound / total weight( 0.879 gm)]*100 A) 1.945*10^(-3) moles and 0.284 gm of CaCl2*2H2O were reacted. B) 3.22*10^(-3) moles and 0.595 gm of K2C2O4*H2O were reacted. C) 0.311 gm of K2C2O4*H2O remain unreacted. D) CaCl2*2H2O = 32.309% and K2C2O4*H2O = 67.69% by mass in the mixture.
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