A 0.972g sample of CaCl2.2H2O/K2C2O4.H2O solid salt mixture is dissolved in 150
ID: 634735 • Letter: A
Question
A 0.972g sample of CaCl2.2H2O/K2C2O4.H2O 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 0.375g the limiting reactant in the salt mixture was later determined to be CaCl2.2H2O
A. What is the % by mass of CaCl2.2H2O in the salt mixture
B. how many grams of the excess reactant K2C2O4.H2O reacted in the mixture
C. How many grams of the K2C2O4. H2O in the salt mixture remain unreacted
Explanation / Answer
Write a balanced equation: Do not include the water of crystallisation in the equation.
CaCl2(aq) + K2C2O4(aq) ? 2 KCl(aq) + CaC2O4(s)
1mol CaCl2 reacts with 1mol K2C2O4 to produce 1mol CaC2O4 (ignore the KCl)
Molar mass CaC2O4 = 128.097 g/mol, anhydrous (146.112 g/mol,monohydrate) Because your precipitate of CaC2O4 was airdried, we have to work with the monohydrate product.
0.375g = 0.375/146.112= 2.567*10^-3 mol monohydrate, which would also be 2.567*10^-3 anhydrous CaC2O4
If we accept that the limiting reactant was the CaCl2, then we can say that 2.567*10^-3 mol K2C2O4 reacted. This would also refer to the moles of the hydrated salt.
We can also say that 2.567*10^-3 mol CaCl2 reacted. This would also refer to the moles of the hydrated salt.
Molar mass K2C2O4.H2O = 39.098*2 + 12.011*2 + 15.99*4 + 1.008*2 + 15.99*1 = 184.184g/mol .
2.567*10^-3*184.184 = 0.473g K2C2O4.H2O reacted ANSWER to Q(a)
Molar mass CaCl2.2H2O = 40.078 + 35.453*2 + 1.008*4 + 15.99*2 =146.996g/mol
2.567*10^-3* 146.996 = 0.377g CaCl2.2H2O reacted
Because the CaCl2.2H2O is the limiting reagent,the 0.377g reacted totally and reacted with 0.473g K2C2O4.H2O , the total mass reacted was 0.850g
You started with 0.972 grams mixture and 0.850g was reacted, leaving 0.122g K2C2O4 unreacted.
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