Question

A student starts with   4.586 g   of unknown mixture of Na₂CO₃ and CaCl₂•2H₂O. They collect   1.356 g   of precipitate. They identified Na₂CO₃ as their limiting reagent.

How many grams of CaCl₂•2H₂O did the unknown mixture originally contain?

ive tried 3.230 and 2.898 but were both wrong.

thanks

Answer 1

Na2CO3 (aq) + CaCl2 .2H2O(aq) ----------------> 2NaCl (aq) + CaCO3 (s)

It is given that mixture of Na₂CO₃ and CaCl₂•2H₂O weighs ------------- 4.586g

Let the amount of Na2CO3 ------------------- x g

the amount of CaCl2. 2H2O ---------------------------- (4.586-x)g

From the balanced equation, we get the information that:

106g of Na2CO3 is reacting with------ 147g of CaCl2. 2H2O and forms---------- 100g of CaCO3

Then ?g of Na2CO3 we get-------------------------------------------------------------------------- 1.356 g of CaCO3

= 1.356g X 106g / 100g

= 1.437g of Na2CO3.

Hence the amount of Na2CO3 in the given mixture x = 1.437g

The amount of CaCl₂•2H₂O in the given unknown mixture= 4.586 g - 1.437g

                                                                                          = 3.149g