Determine the limiting reagent.
Determine how much of the excess reagent is present and how much of (it will be all of it) the limiting reagent reacted.
Using the limiting reagent, calculate the amount of heat generated from the -116 kJ. That will be q.
Then
q = mass x specific heat x (Tfinal - Tinitial). Solve for Tfinal. Show any work if you need additional assistance.
2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(l)
H = -118 kJ
Calculate the heat when 112.4 mL of 0.500 M HCl is mixed with 300.0 mL of 0.515 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 412.4 g and a specific heat capacity of 4.18 J/°C·g, calculate the final temperature of the mixture.
in C
9 answers
i know that i am slow here at chem but how do i find out the limiting reactant... i think it is the HCl but i am not sure. then how to do the rest of the problem. i have a few others like this problem. if you have time can you help me work it step by step?
Determine the mols HCl you have. That is Liters x molarity. Determines mols Ba(OH)2 you have. That is liters x molarity. The surest way of doing this, I think, is to calculate the mols of BaCl2 formed. I will do the HCl one.
mols HCl = L x M = 0.1124 L x 0.500 M = 0.05620 mols HCl. Now how many mols BaCl2 will that form? It will be
mols BaCl2 formed = 0.05620 mols HCl x (1 mol BaCl2/2 mols HCl) = 0.2810 mols BaCl2.
Do the same thing for Ba(OH)2 and convert that to mols BaCl2. The limiting reagent will be the one that produces the FEWER mols BaCl2.
mols HCl = L x M = 0.1124 L x 0.500 M = 0.05620 mols HCl. Now how many mols BaCl2 will that form? It will be
mols BaCl2 formed = 0.05620 mols HCl x (1 mol BaCl2/2 mols HCl) = 0.2810 mols BaCl2.
Do the same thing for Ba(OH)2 and convert that to mols BaCl2. The limiting reagent will be the one that produces the FEWER mols BaCl2.
for the mols of Ba(OH)2 i got .1545 so in this reaction the limiting reagent is the Ba(OH)2? is that correct?
0.1545 mols Ba(OH)2 is correct. How many mols BaCl2 will each reactant (HCl and Ba(OH)2) produce. The smaller value will tell you which is the limiting reagent.
OH i see i didn't continue with the equation in that case .. Hcl will produce .2810mols and Ba(OH)2 will still produce the .1545mol b/c ration is one to one, correct?
I think you meant 0.02810 mols BaCl2 for HCl since 1/2 x 0.05620 = 0.02810.
So HCl will produce 0.02810 mols BaCl2. The Ba(OH)2 will produce 0.1545 mols BaCl2. The smaller number is 0.02810 meaning HCl is the limiting reagent.
Now you know 0.02810 mols HCl will react. What is delta H for the reaction if 0.02810 mols HCl react. The problem tells you it is -118 if 2 mols react? That is substituted for q in
q = mass x specific heat x (Tfinal - Tinitial)
So HCl will produce 0.02810 mols BaCl2. The Ba(OH)2 will produce 0.1545 mols BaCl2. The smaller number is 0.02810 meaning HCl is the limiting reagent.
Now you know 0.02810 mols HCl will react. What is delta H for the reaction if 0.02810 mols HCl react. The problem tells you it is -118 if 2 mols react? That is substituted for q in
q = mass x specific heat x (Tfinal - Tinitial)
all you have to do then is divide the -118 by 2 and then multiple that answer by .02810?
so q would then be -1.6579?
so q would then be -1.6579?
I really don't know.
If the -118 is kJ/mol, then, yes, you divide the -118 by 2 and multiply by 0.02810. But if the -118 kJ is the heat given off by the reaction (not kJ/mol but kJ/reaction), then -118 x 0.02810 is the way to do it. I don't know if your reaction is kJ/mol or kJ/rxn.
If the -118 is kJ/mol, then, yes, you divide the -118 by 2 and multiply by 0.02810. But if the -118 kJ is the heat given off by the reaction (not kJ/mol but kJ/reaction), then -118 x 0.02810 is the way to do it. I don't know if your reaction is kJ/mol or kJ/rxn.
2 answers