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

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

these nuts