contestada

10. When 50.0 ml of 0.200 M AgNO3 and 50.0 ml of 0.100 M CaCl2, both at 25.0°C, are reacted in a coffee-cup calorimeter, the temperature of the reacting mixture increases to 26.0°C. Calculate ∆H in kJ per mole of AgCl produced. Assume the density of the solution is 1.05 g/ml and the specific heat capacity of the solution 4.20 J/g°C.AgNO3(aq) + HCl(aq) à AgCl(s) + HNO3(aq)

Respuesta :

dsdrajlin

Answer:

ΔH° = 840 kJ/mol

Explanation:

Let's consider the following balanced reaction:

2 AgNO₃(aq) + CaCl₂(aq) ⇄ 2 AgCl(s) + Ca(NO₃)₂

Then, we need to know the moles of both reactants:

AgNO₃: n = 0.200 mol/L × 0.0500 L = 0.0100 mol

CaCl₂:   n = 0.100 mol/L × 0.0500 L = 0.00500 mol

According to the balanced equation we need 2 moles of AgNO₃ per each mole of CaCl₂, and this coincides with the experimental data, so there is no limiting reactant. Let's use AgNO₃ to find out how many moles of AgCl are produced.

[tex]0.0100molAgNO_{3}.\frac{2molAgCl}{2molAgNO_{3}} =0.0100molAgCl[/tex]

Now, we can calculate the total amount of heat released using the following expression:

Q = c × m × ΔT

where,

c is the heat capacity of the solution

m is the mass of the solution

ΔT is the change in temperature (26.0 °C - 25.0°C = 1.00 °C)

Since the volume is 100.0 mL (50.0 mL + 50.0 mL) and the density is 1.05 g/mL, we can calculate the mass of the solution like:

m = 1.05 g/mL × 100.0 mL = 105 g

Then,

[tex]Q=c \times m \times \Delta T=4.20 J/g.\° C \times 105 g \times 1.00\° C=8.40 \times 10^{3} J=8.40kJ[/tex]

Finally,

[tex]\Delta H \° = \frac{8.40kJ}{0.0100mol} =840kJ/mol[/tex]

Otras preguntas