Should the Town Buy NaCl or CaCl2 for De-icing?”

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The discussion centers on the cost-effectiveness of using NaCl versus CaCl2 for de-icing in a small town. NaCl is priced at $0.36/kg, and participants calculate the moles produced from each compound, finding that 1 kg of NaCl yields approximately 17.12 moles of ions, while 1 kg of CaCl2 yields 9.0 moles of Ca and 18.0 moles of Cl. The cost-effectiveness analysis suggests that the town should pay no more than 0.52 to 1.05 times the price of NaCl for CaCl2, depending on whether the focus is on total ion contribution or just chloride ions. The discussion also notes that the hydroscopic nature of CaCl2 complicates its practical use. Overall, the analysis emphasizes the importance of considering both cost and de-icing effectiveness when making the purchasing decision.
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Homework Statement




"De-icing salt" is used to melt snow and ice on streets. the highway department of a small town is decideing whether to by NaCl or CaCl2 for the job. The town can obtain NaCl for $0.36/kg. What is the maximum the town should pay for CaCl2 to be cost effective? in kg



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The Attempt at a Solution




I am sure this is very simple but I am not sure what to do with it...
 
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Do you know about moles and molar mass of compounds?
Do you know how to work out how much NaCl there is in a kg?

You have posted 3 questions which are basically the same thing - they all need this step.
 
I am really not sure...i am pretty new to chemistry and not very good at it, can i get some help
 
Can you work out how many moles are in 1 kg of NaCl and CaCl2? What part are you confused on?
 
i worked out .05844 mol in 1 kg for NaCL
and .11098 mol in 1 kg of CaCl2

is this correct
 
No, molar mass of NaCl is 58.4g, CaCl2 is 111g
but for every mole of CaCl2 you get twice as much Cl as you do from NaCl

So 1kg of NaCl gets you 17.12moles of Na and Cl, 1kg of CaCl2 is 9.0 moles of Ca and 18.0moles of Cl

It isn't clear from the question if you supposed to take into account the de-icing effect of the Cl ion so the answer is either.
That you should pay 9/17.2 = 0.52 times as much for CaCl2
or if you are only concerned about Cl then you should pay 18/17.2 = 1.05 times as much for CaCl2


(Actually it's a very stupid example because CaCl2 is so hydroscopic it will almost certainly not be 111g/mole - but since this is an intro chemistry question you can ignore this)
 
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