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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Unknown Username on December 01, 2008, 09:01:24 PM

Title: Moles
Post by: Unknown Username on December 01, 2008, 09:01:24 PM

Hi!

When 12.0 g potassium nitrate (KNO3)  is dissolved in 100 cm3 of water, the temperature drops by 4.20 °C.
     
       Relative atomic mass: K=39, N=14; O=16
       Specific heat capacity of water = 4.2 J g–1 K–1
       Heat change = M x C x ΔT
       1 cm3 of soltuion = 1 g

(i)    Using the above information calculate EACH of the following:

       a)  The number of  KNO3 used in the experiment
            _____________________________________

        b)   The heat change for the reaction
             _____________________________________
        c)   The enthalpy change in kJ in mol –1 for the reaction
              _____________________________________
ii) State ONE assumption you made in your calculation.
              _____________________________________ 

May I, kindly, receive some assistance with this question, please?

Title: Re: Moles
Post by: Astrokel on December 02, 2008, 05:41:04 AM

You have to show some work!

i) I believe you can do (a) and equation has already given for (b), also for (c) look at the units.

ii) Hint: A famous law in science

Title: Limiting Reagent
Post by: Accord on December 02, 2008, 04:10:45 PM

How to find out the Limiting Reagent in an equation?
 

Title: Re: Moles
Post by: Borek on December 02, 2008, 04:37:43 PM

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents

Title: Re: Moles
Post by: Unknown Username on January 05, 2009, 03:52:22 PM

12g KNO3
RAM KNO3= 39+14+9+(16x3)
              =101
#moles = mass present  = 12  = 0.1188mol
                  RAM            101
 
b) /\ C= M x C x /\ T
         = 12g x 4.2Jg-1K-1 x 4.20C
         = 211.68J

Can't do the rest.
 

Title: Re: Moles
Post by: Astrokel on January 05, 2009, 04:14:34 PM

Quote

b) /\ C= M x C x /\ T
         = 12g x 4.2Jg-1K-1 x 4.20C
         = 211.68J

You are close there but M represents mass of solution!

Part c is pretty easy. Look at the units again, how is it linked to your previous parts? Also determine if it is an endo or exo reaction for appropriate +/- sign.