A student starts with4 586 g ofunknown mixture of Na 2 CO 3 andCaCl 2 ...

A student starts withÂ Â Â 4.586 gÂ Â of unknown mixture of Na2CO3 and CaCl2â€¢2H2O. They collectÂ Â Â 1.356 gÂ Â of precipitate. They identified...

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Chemistry

A student starts withÂ Â Â 4.586 gÂ Â ofunknown mixture of Na₂CO₃ andCaCl₂â€¢2H₂O. TheycollectÂ Â Â 1.356 gÂ Â of precipitate. Theyidentified Na₂CO₃ as their limitingreagent.
How many grams of CaCl₂â€¢2H₂O did theunknown mixture originally contain?
ive tried 3.230 and 2.898 but were both wrong.
thanks

Answer & Explanation Solved by verified expert

4.3 Ratings (862 Votes)

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

It is given that mixture of Na₂CO₃ and CaCl₂â€¢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 CaCl₂â€¢2H₂O in the given unknown mixture= 4.586 g - 1.437g

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3.149g

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