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Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 ∘C to 27.7 ∘C.
a) Before mixing, how many grams of Cu are present in the solution of CuSO4?
b) Predict the identity of the precipitate in the reaction.
c) Write complete equation for the reaction that occurs when the two solutions are mixed.
d) Write net ionic equation for the reaction that occurs when the two solutions are mixed.
e) From the calorimetric data, calculate ΔH for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, and that the specific heat and density of the solution after mixing are the same as that of pure water.

Respuesta :

Explanation:

Molarity of copper sulfate solution = 1.00 M

Volume of the copper sulfate solution  = 50.0 mL = 0.050 L

Moles of copper sulfate = n

[tex]1.00M=\frac{n}{0.050 L}[/tex]

n = 0.050 L × 1.00 M= 0.050 mol

1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :

[tex]1\times 0.050 mol=0.050 mol[/tex] of copper

a) Mass of 0.050 moles of copper = 0.050 mol × 63.5 g/mol =3.175 g

b) The identity of the compound which formed after the reaction is copper hydroxide.

c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

[tex]CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)[/tex]

d) [tex]CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)[/tex]..[1]

[tex]KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)[/tex]..[2]

[tex]Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)[/tex]

Common ion both sides are removed. The net ionic equation is given as:

[tex]Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)[/tex]

e) Volume of solution after mixing of both solution,V= 50 mL + 50ml = 100 mL

Mass of final solution ,m= 1 mL

Density of solution ,d= 1 g/mL (same as pure water)

[tex]m=d\time V=1 g/ml\times 100 mL = 100 g[/tex]

Heat capacity of the solution = c = 4.186 J/g°C (same as pure water)

Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

[tex]Q=mc\Delta T[/tex]

[tex]Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ[/tex]

Enthalpy of the reaction = ΔH = [tex]\frac{Q}{\text{Moles of copper}}[/tex]

ΔH = [tex]\frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol[/tex]

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

pstnonsonjoku

The enthalpy change for the reaction is -51.8 KJ/mol.

The equation of the reaction is;

CuSO4(aq) + 2KOH(aq) -----> Cu(OH)2(s) + K2SO4(aq)

The complete ionic equation of the reaction is;

Cu^2+(aq) + SO4^2-(aq) + 2K^+(aq) + 2OH^-(aq) -----> Cu(OH)2(s) + 2K^+(aq) +  SO4^2-(aq)

Net ionic equation = Cu^2+(aq) + 2OH^-(aq) -----> Cu(OH)2(s)

Number of moles of CuSO4 = 50/1000 L × 1.00 M = 0.05 moles

Number of moles of  KOH =  50/1000 L ×  2.00 M = 0.1 moles

 CuSO4 is the limiting reactant here

Since the specific heat and density of the solution after mixing are the same as that of pure water;

c = 4.18 J/g/∘C

d = 1 g/ml

Total volume of solution =  100.0 mL

Total mass of solution = 100.0 g

From;

ΔH =- mcθ

m = mass of water

c = specific heat capacity of water

θ = temperature rise = 27.7 ∘C - 21.5 ∘C = 6.2∘C

ΔH =  -(100.0 g ×  4.18 J/g/∘C ×  6.2∘C )/ 0.05 moles

ΔH = -51.8 KJ/mol

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