Homework Answers
Volume of HCl = 100 ml
Concentration of HCl = 0.4 M
Moles of HCl = 100 x 0.4 / 1000 = 0.04 Moles
Volume of NH4OH = 150 ml
Concentration of NH4OH = 0.3 M
Moles of NH4OH = 150 x 0.3 / 1000 = 0.045 Moles
Among two reactants, NH4OH is excess. Hence HCl is limiting reagent
Moles of NH4Cl produced = 0.04 Moles
Mass of the solution = 100 ml + 150 ml = 250 ml
Temperature raised = 26.07∙oC - 23.92 ∙oC = 2.15 ∙oC
we can calculate the heat released from the following formula
Q = mc∆T
Q = heat energy (Joules, J), m = mass of a substance (g)
c = specific heat (units J/g∙oC), ∆ is a symbol meaning "the change in"
∆T = change in temperature (oC Celcius)
Q = 250 gx 4.184 J/g∙oC x 2.15 ∙oC
Q = 2248.9 Joules or 2.248 K J
Enthalpy of the reaction = 2.248 / 0.04 = 56.222 KJ
Limiting reagent is HCl.
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