If the reaction of 25.0 mL of 1.0 M HBr and 35.0 mL of 1.0 M KOH produces 1.45 K

Question

If the reaction of 25.0 mL of 1.0 M HBr and 35.0 mL of 1.0 M KOH produces 1.45 KJ of heat (called q in this experiment), what is the delta H of HBR + KOH(aq) per mole of water formed? Show your work, include units, and use the correct number of significant figures. Given the heat of the following reaction: Calculate the amount of heat produced (called q in experiment) by reacting 50.0 ml of 2.0 M NH4CI with 50.0 mL of 2.0 M NaoH. Show your work and include units. 3. Using the q calculated in 02 and the other information given in that problem, calculate what the delta T would be in that reaction Assume the specific heat capacity of the solutions is 4.184 J/g degree C. Show your work. (Such a small change in temperature would be Mat for us to detect on our thermometers. That's why no students earned out the reaction shown in #2.)

Explanation / Answer

There are multiple questions here . i am allowed to answer only 1 at a time. I will answer 1st one for you. Please ask other as different question.
1.
moles of HBr = M*V = 1 M * 0.025 L = 0.025 moles
moles of KOH = M*V = 1 M * 0.035 L = 0.035 mol

HBr (aq)+ KOH (aq) ---> KBr (aq) + H2O (l)

clealr HBr is limiting reagent
So,
when 0.025 mol of each reacts, heat produced = 1.45 KJ
when 1 mol of each will react, heat produced = 1.45 / 0.025 = 58 KJ

so,
delta H of reaction = -58 KJ/mol
Answer: -58 KJ/mol

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---