Mg + 2HC2H3O2 ==> Mg(C2H3O2)2 + H2
mols Mg = 0.7/24.3 = about 0.0288
You ask for two things; i.e., you ask for enthalpy of Mg AND initial temperature. Here is how you do Ti. Assuming specific heat AND density vinegar are the same as a solution of water then
q = mass soln x specific heat soln x (Tfinal-Tinitial)
-123400 J = 60 g x 4.184 J/oC*g x (31.8 - Ti)
Substitute and solve for Ti.
For the enthalpy, dH is -123.4 kJ for 0.0288 mols Mg.
So -123.4 kJ x (1 mol/0.0288 mol Mg) = ? kJ/mol
Calorimeter lab. What is the change in enthalpy for magnesium after the reaction with vinegar?
vinegar is 60ml
magnesium is 0.7g
T1 = ?
T2 = 31.8°C
find the initial temperature if 123.4 KJ energy has been released.
THIS IS ALL THE INFORMATION THAT I GOT FROM MY TEACHER I SWEAR.
1 answer
2 answers