I need help with this problem
Predict the final temperature of the solution when you mix 120.0 mL of 0.50 M NaOH with 30.0 mL of 0.75 M HCN in a coffee cup calorimeter at 25 degrees C (initial temperature), using the following thermomechanical equation:
NaOH (aq) + HCN (aq) ---> NaCN (aq) + H20 (l)
ΔH = -45 J
(You may assume that all solutions have a density of 1.00 g/mL, and that the specific heat of the combined solutions is 4.18 J/g-K)
I tried to do this, but I'm not sure which formula(s) to use and how to plug in the information given.
thanks in advance
Predict the final temperature of the solution when you mix 120.0 mL of 0.50 M NaOH with 30.0 mL of 0.75 M HCN in a coffee cup calorimeter at 25 degrees C (initial temperature), using the following thermomechanical equation:
NaOH (aq) + HCN (aq) ---> NaCN (aq) + H20 (l)
ΔH = -45 J
(You may assume that all solutions have a density of 1.00 g/mL, and that the specific heat of the combined solutions is 4.18 J/g-K)
I tried to do this, but I'm not sure which formula(s) to use and how to plug in the information given.
thanks in advance
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moles of each substance:
NaOH ---> (0.50 ml/L) (0.120 L) = 0.060 mol
HCl ---> (0.75 mol/L) (0.030 L) = 0.0225 mol
The HCl is the limiting reagent. 0.0225 of the HCl reacts with 0.0225 mol of NaOH. We ignore the remaining amount of NaOH.
How much heat energy was released?
45000 times 0.0225 = 1012.5 J
You wrote -45 J, it's -45 kJ
Determine temperature increase:
1012.5 J = (150 g) (x) (4.184 J/g C)
x = 1.6 C
final temp = 26.6 C
NaOH ---> (0.50 ml/L) (0.120 L) = 0.060 mol
HCl ---> (0.75 mol/L) (0.030 L) = 0.0225 mol
The HCl is the limiting reagent. 0.0225 of the HCl reacts with 0.0225 mol of NaOH. We ignore the remaining amount of NaOH.
How much heat energy was released?
45000 times 0.0225 = 1012.5 J
You wrote -45 J, it's -45 kJ
Determine temperature increase:
1012.5 J = (150 g) (x) (4.184 J/g C)
x = 1.6 C
final temp = 26.6 C