Respuesta :
Explanation:
The reaction equation will be as follows.
[tex]Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)[/tex]
So, according to this equation, 1 mole [tex]Ca(OH)_{2}[/tex] = 2 mol HBr = 1 mol [tex]CaBr_{2}[/tex]
Therefore, calculate the number of moles of calcium hydroxide as follows.
No. of moles of [tex]Ca(OH)_{2}[/tex] = [tex]V \times Molarity[/tex]
= [tex]50 \times 0.6[/tex]
= 30 mmol
Similarly, calculate the number of moles of HBr as follows.
No. of moles of HBr = [tex]M \times V[/tex]
= [tex]50 \times 0.6[/tex]
= 30 mmol
This means that the limiting reactant is HBr.
So, no. of moles of [tex]CaBr_{2}[/tex] = [tex]30 \times \frac{1}{2}[/tex]
= 15 mmol
Hence, calculate the amount of heat released as follows.
Heat released in the reaction(q) = [tex]m \times s \times \Delta T[/tex]
as, m = mass of solution
and, Density = [tex]\frac{mass}{volume}[/tex]
or, mass = Density × Volume
= 1.08 g/ml \times (50 + 50) ml
= 108 g
where, s = specific heat of solution = 4.18 j/g.k
and, change in temperature [tex]\Delta T[/tex] = [tex](26 - 23)^{o}C[/tex]
= [tex]3 ^{o}C[/tex]
Hence, the heat released will be as follows.
q = [tex]m \times s \times \Delta T[/tex]
q = [tex]108 \times 4.18 \times 3^{o}C[/tex]
= 1354.32 joule
or, = 1.354 kJ (as 1 kJ = 1000 J)
Also, [tex]\Delta H_{rxn}[/tex] = [tex]\frac{-q}{n}[/tex]
= [tex]\frac{-1.354}{15 \times 10^{-3}}[/tex]
= -90.267 kJ/mol
Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.
