# Writing net ionic equation and stoichiometry

A 0.879-g sample of a CaCl2.2H2O/K2C2O4H2O solid salt mixture is dissolved

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in approximately 150mL of deionized? rest of Prblm: 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?

### Lkely the precipitate will be

Lkely the precipitate will be CaC2O4 and if so reaction is

CaCl2(aq) + K2C2O4(aq) ---> CaC2O4(s) + 2KCl(aq)

Ca2+(aq) + 2Cl-(aq) + 2K+(aq) + C2O42-(aq) ---> CaC2O4(s) + 2Cl-(aq) + 2K+(aq) which becomes Ca2+(aq) + C2O42-(aq) ---> CaC2O4(s)

CaCl2(aq) + K2C2O4(aq) ---> CaC2O4(s) + 2KCl(aq)

1mole   +  1 moles   ---->  1 mole

Find moles masses of the  moles values and then find the mass of CaCl2 needed to make 0.284g of CaC2O4(s)

Subtract 0.284g from 0.879g to find mass of other salt