A 100.0 ml sample of 0.300 M NaOH is mixed with a 100.0 ml sample of 0.300 M HNO3 in a coffee cup calorimeter. Both solutions were initially at 35.0 degrees celcius; the temperature of the resulting solution was recorded at 37.0 degrees celcius. Determine the delta H (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HNO3. Assume no heat is lost to the calorimeter or the surroundings; and the density and heat capacity of the resulting solution are the same as water.
SO I DID THIS PROCEDURE
0.1 L x 0.300 M NaOH = 0.0300 moles NaOH.
0.1 L x 0.300 M HNO3 = 0.0300 moles HNO3.
Mass water (200 mL) x specific heat water x (Tfinal-Tinitial) = q in joules.
18(200 mL)*4.18*(37-35)=30096 joules
THEN DIVIDE IT BY 0.03 = 1003200 JOULES/MOLE
THEN I TURNED IT INTO KILOJOULES WHICH I THEN GOT 1003.2 KILOJOULES/MOL.
IS THIS CORRECT?
1 answer
OH I THINK I MADE A MISTAKE. I GOT 55.7 AS MY ANSWER NOW
1 answer