When 25.0 mL of 0.25 mol/L LiOH and 25.0 mL of 0.25 mol/L HCl are mixed together, the temperature warms 15.8°C. What is the molar enthalpy of neutralization for LiOH?  I attached my answer but am unsure.

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1. When 25.0 mL of 0.25 mol/L LiOH and 25.0 mL of 0.25 mol/L HCl are mixed together, the temperature warms 15.8°C. What is the molar enthalpy of neutralization for LiOH? 

I attached my answer but am unsure. 

mHci = 25.0mL = 25.0g
CHCI = 0.25mol/L
V LIOH = 25.0mL = 0.025L
CLIOH = 0.25mol/L
CH,0 = 4.184J/g°c
AT = 15.8°C
AH°,
= ?
пеut
V
water
+V HCI
HCI
= V
LIOH
= 25.0mL +25.0mL
50mL
mH,0 = 50g
Owater = MCAT
= (50g)(4.184J/g°C)(15.8°C)
= 3305.36J
AH°
Qwater
system
3305.36J
NLIOH = (CLiOH)(V Lioh)
= (0.25mol/L)(0.025L)
= 0.0063mol
AH°
AH°
system
пеut
-3305.36/
0.0063mol
- 524660J/mol
- 525kJ/mol
Transcribed Image Text:mHci = 25.0mL = 25.0g CHCI = 0.25mol/L V LIOH = 25.0mL = 0.025L CLIOH = 0.25mol/L CH,0 = 4.184J/g°c AT = 15.8°C AH°, = ? пеut V water +V HCI HCI = V LIOH = 25.0mL +25.0mL 50mL mH,0 = 50g Owater = MCAT = (50g)(4.184J/g°C)(15.8°C) = 3305.36J AH° Qwater system 3305.36J NLIOH = (CLiOH)(V Lioh) = (0.25mol/L)(0.025L) = 0.0063mol AH° AH° system пеut -3305.36/ 0.0063mol - 524660J/mol - 525kJ/mol
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