Answer:
[tex]463.98~KJ/mol[/tex]
Explanation:
We have to start with the balance reaction:
[tex]Mg_(_s_)+~2HCl~_(_a_q_)->MgCl_2_(_a_q_)~+~H_2_(_g_)[/tex]
Now we have to keep in mind the heat equation:
[tex]Q=m*Cp*[/tex]ΔT
Where:
Q=heat
m= Mass
ΔT= Tfinal-Tinitial
Cp= 4.18 [tex]\frac{J}{g^{\circ}C}[/tex]
Calculation of mass: We have to use the density of water [tex]1~\frac{g}{mL}[/tex]
[tex]100~mL\frac{1~g}{1~mL} =100~ g[/tex]
Calculation of ΔT:
ΔT = 34.1-23.0= 11.1 ºC
Now, we can calculate Q:
[tex]Q=100~g*4.18~\frac{J}{g^{\circ}C}~*11.1^{\circ}C=4639.8~J[/tex]
[tex]4639.8~J~=~4.63KJ[/tex]
To get the final answer we must find the moles of [tex]Mg^+^2[/tex], so:
[tex]0.243~g~Mg~\frac{1~mol~Mg}{24.3~g~Mg}=0.01~mol~Mg[/tex]
With the heat and the moles we have calculate the KJ/mol:
[tex]\frac{4.63~KJ}{0.01~mol~Mg}=~463.98~KJ/mol[/tex]
When 0.243 g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant-pressure calorimeter, leading to a temperature increase from 23.0 °C to 34.1 °C, the enthalpy of the reaction is -464 kJ/mol.
When 0.243 g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant-pressure calorimeter, the following reaction occurs:
Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
If the temperature of the solution increases from 23.0 °C to 34.1 °C as a result of this reaction, calculate ΔH in kJ/ mol of Mg. Assume that the solution has a specific heat of 4.18 J/g°C.
First, we will assume the density of the solution is the same as water (1 g/mL). The mass corresponding to 100 mL of solution is:
[tex]100 mL \times \frac{1g}{mL} = 100 g[/tex]
We can calculate the heat absorbed by the solution (Qs) using the following expression.
[tex]Qs = c \times m \times \Delta T = \frac{4.18J}{g.\° C} \times 100g \times (34.1\° C - 23.0 \° C) \times \frac{1kJ}{1000J} = 4.64 kJ[/tex]
where,
According to the law of conservation of energy, the sum of the heat absorbed by the solution (Qs) and the heat released by the reaction (Qr) is zero.
[tex]Qs + Qr = 0\\\\Qr = -Qs = -4.64 kJ[/tex]
The molar mass of Mg is 24.3 g/mol. The moles (n) corresponding to 0.243 g of Mg are:
[tex]0.243 g \times \frac{1mol}{24.3g} = 0.0100 mol[/tex]
Finally, we will calculate the enthalpy of the reaction (∆H) using the following expression.
[tex]\Delta H = \frac{Qr}{n} = \frac{-4.64kJ}{0.0100 mol} = -464 kJ/mol[/tex]
When 0.243 g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant-pressure calorimeter, leading to a temperature increase from 23.0 °C to 34.1 °C, the enthalpy of the reaction is -464 kJ/mol.
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