A 52.19 mL sample of 3.02 M CuSO4 reacts with 54.48 mL of 2.87 M KOH in a coffee cup calorimeter to form Cu(OH)2(s) and K2SO4 (aq). The initial temperature is 18.5°C and the final temperature is 44.9°C. Assume that the mixture has the same specific heat as water and the same density as water. Specific heat of water is 4.184 J/g-K. a.) Write a chemical equation for the reaction. b.) How many moles of CuSO4 reacted? c.) How many moles of KOH reacted? d.) Which reactant is the limiting reagent? e.) How much heat, in J, was transferred to the solution? f.) How much heat, in kJ, was transferred by the reaction? g.) Calculate the heat transferred by the reaction in kJ/ mol (use the limiting reagent moles) h.) Is the reaction exothermic or endothermic?

a) see the chemical equation for the reaction below

b)  0.1576 Moles

c) 0.1563 Moles  

d) The limiting reagent is KOH which is present in less amount.

h) it is an Endothermic Reaction

Explanation:

a) CuSO₄ + KOH > Cu(OH)₂ (s)+ K₂2SO₄ (aq)

b) Moles of CuSO₄ reacted = 52.19 x 3.02 / 1000 = 0.1576 Moles

c) Moles of KOH reacted = 54.48 x 2.87 /1000 = 0.1563 Moles

to get the heat transferred,

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙°C), ∆ is a symbol meaning "the change in"

∆T = change in temperature (°C Celcius)

m = 54.48 + 52.19 = 106.67 ml = 106.67 gm

Q = 106.67 gm x 4.184 J/g∙o°C x ( 44.9°C - 18.5°C)

Q = 11782.51 Joules = 11.782 Kilo Joules

Hence 11.782 Kilo Joules of heat was transferred into the solution