how many grams of CaCl2 must dissolve to release 192 KJ of heat? answer is 258 g but i need to know the process! help please! quiz tomorow!

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- anonymous

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- anonymous

\[CaCl _{2}\rightarrow Ca _{aq.}+2Cl _{aq.} ; DeltaHsoln=-82.8kj/mol\]

- anonymous

equation was given with problem

- anonymous

ok using you balanced thermo equation we can pull out the conversion factor
1mole cacl2=82.8 kj
Now we factor lable this bad boy,
\[192 kj (\frac{1 mol CaCl_2}{82.8 kj})(\frac{111 g CaCl_2}{1 mole CaCl_2})=\]

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## More answers

- anonymous

im sorry im still kinda lost what do you do after its set up like that?

- anonymous

take 192 multiply it to the top of the fraction then divide it by the bottom of the raction till you get to the end and you will have your answer.

- anonymous

okay i understand a bit better and the 111 came from the table of elements added up ca and cl2 right?

- anonymous

Yea but i'm guessing there is a difference between the table i use compaired to the one the answer came from or a rounding error or something because I got 257.39 kj i think it was. This also could be a case were carring numbers gives us a slightly different answer. However thats how it should be set up you can look at the molar mass and see if its the same or slightly different off your peroidic table.

- anonymous

yea i got it thanks!

- anonymous

No problem good luck on your quiz tomorrow

- anonymous

yea lol im going to need it do you think you can help me with one more?

- anonymous

I can try

- anonymous

thanks sorry for the wait its the same eq. but its asking for heat released if you prepeare 2.75L of 1.75 M [CaCl2 solution

- anonymous

and the answer was 285 Kj

- anonymous

\[2.75 L (\frac{1.75 mol}{1 L})(\frac{82.8 kj}{1 mole})=\]
It should look something like that but this one isn't working out and i'm getting tired. Is there anything that you left out in the question?

- anonymous

no thats all there was in the question, and thanks for trying lol ill ask a classmate tomorrow morning lol i still have to study for my math and physics test too so i cant put to much time on thsi but thanks for the help i appreciate it :)

- anonymous

np and sorry i couldn't figure out this last one

- anonymous

thats okay, lucky for me i know all the smarties in the class so they will help me out so not to worry besides i have an hour in the morining to figure this out before i get to school so np :)

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