196 4. A 0.879-g sample of a Ca 2HO/K,CO HOsolid salt mixture is dissolved in 15

Question

196 4. A 0.879-g sample of a Ca 2HO/K,CO HOsolid 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 0.284 g. The limiting reactant in the salt mixture was later determined to be CaCl, 2H,O. Write the molecular form of the equation for the reaction. a. b. Write the net equation for the reaction. c. How many moles and grams of CaCl, 2H 0 reacted in the reaction mixture? d. How many moles and grams of the excess reactant, K.c.o,1,0, reacted in the mixture? How many grams of the KC,0..,0 in the salt mixture remain unreacted (in ecess)? e. f. What is the percent by mass of each salt in the mixture?

Explanation / Answer

a.)

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

b)

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

ionic equation : Ca^2+(aq)+ 2Cl^-(aq) + 2K^+(aq)+ C2O4^2-(aq) ----> CaC2O4(s) + 2K^+(aq) + 2Cl^-(aq)

cancel the spectator ions on both sides.

net equation : Ca^2+(aq) + C2O4^2-(aq) ----> CaC2O4(s)

c)

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

   1 mol CaCl2 = 1 mol K2C2O4 = 1 mol CaC2O4

from the given data ,

No of mole of CaC2O4 formed = w/mwt = 0.284/128.097 = 0.002217 mol

so that,

No of mol of CaCl2*2H2O reacted = 0.002217 mol

Amount of CaCl2*2H2O reacted = n*Mwt

                              = 0.002217*147.015

                               = 0.326 g

d)

According to the question

We know that,

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)

then,

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

then,

1.945*10^-3 moles and 0.284 gm of CaCl2*2H2O were reacted.

Then

3.22*10^-3moles and 0.595 gm of K2C2O4*H2O were reacted

e)

The sample contains both CaCl2.2H2O and K2C2O4.H2O.

Mass of sample = 0.879 g

The reaction between the two is:

CaCl2 + K2C2O4 ----> CaC2O4 + 2KCl

CaC2O4 gets precipitated.

The mass of precipitate is 0.284 g

So, mass of CaC2O4 = 0.284 g

molar mass of CaC2O4 = 128 g/mol

So, moles of CaC2O4 formed = mass/molar mass = 0.284/128 = 0.0022 mol

As CaCl2.2H2O is limiting reagent, so moles of CaCl2.2H2O used = moles of CaC2O4 formed = 0.0022 mol

molar mass of CaCl2.2H2O = 147 g/mol

So, mass of CaCl2.2H2O taken initially = moles*molar mass = 0.0022*147 = 0.323 g

mass of K2C2O4.H2O taken initially = mass of sample - mass of CaCl2.2H2O taken initially

= 0.879-0.323 = 0.556 g

moles of K2C2O4.H2O reacted = 0.0022 mol

molar mass of K2C2O4.H2O = 184 g/mol

mass of K2C2O4.H2O reacted = 0.0022*184 = 0.405 g

mass of K2C2O4.H2O unreacted = mass of K2C2O4.H2O taken initially - mass of K2C2O4.H2O reacted

= 0.556-0.405 = 0.151 g

Mass of K2C2O4.H2O in the salt mixture remained unreacted = 0.151 g

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