Given the following equations: H3BO3(aq) rightarrow HBO2(aq) + H2O(I) H2B4O2(aq)

Question

Given the following equations: H3BO3(aq) rightarrow HBO2(aq) + H2O(I) H2B4O2(aq) + H2O(I) rightarrow H2O(I) rightarrow 4HBO2(aq) H2B4O2(aq) rightarrow 2B2O3(s) + H2O(I) Find the delta H for the overall reaction: 2H3BO3(aq) rightarrow B2O3(s) + 3H2O(I) A 1.201 g sample of salicylic acid (C2H6O3) has been put into a bomb calorimeter. In raises the temperature of the calorimeter from 23.68 degree C to 29.82 dereeC. If the heat of combustion for salicylic acid is -3.023(103) kJ/mol, what is the heat capacity of the bomb calorimeter? (The molar mass of salicylic acid is 138.1 g/mol) Calculate the amount of heat released in the complete combustion of 8.17 g odf Al to form Al2O3(s) at 25 degree C and I atm: 4Al(s) + 3O2(g) rightarrow 2Al2O3(s) The delta Hf6 for Al2O3(s): -1,676 kJ/mol.

Explanation / Answer

9)

reaction 1 :   H3B03 ----> HB02 + H20

reaction 2 : H2B407 + H20 ----> 4 HB02

reaction 3 : H2B407 ---> 2B203 + H20

final reaction : 2H3B03 ----> B203 + 3H20

from the above reactions

final reaction = 2 x reaction1 - ( 0.5 x reaction2 ) + ( 0.5 x reaction3)

dH rxn = 2 x dHrxn 1 - 0.5 dHrxn2 + 0.5 x dHrxn3

dHrxn = - 2 x 0.02 + 0.5 x 11.3 + 0.5 x 17.5

dHrxn = 14.4 kJ

so the answer is option   D

10)

Heat capacity = heat of combustion / change in temp

Heat capacity = 3.023 x 103 / ( 29.82 - 23.68 )

Heat capacity = 492.3 kJ / C

so the answer is option E

11)

given mass of Al = 8.27 g

moles = mass / molar mass

moles of Al = 8.27 / 27

moles of Al = 0.30

4Al + 302 ----. 2Al203

heat released = dHfo products - dHfo reactants

as reactants are in their elemental form dHfo reactants = 0

so

heat released = dHfo products

heat released = moles of Al203 x dHfo Al203
from the reaction

moles of Al203 = moles of Al / 2

so

moles of AL203 = 0.30/2

moles of Al203 = 0.15

given

dHfo Al203 = -1676 kJ / mol

Heat released = moles x dHfo

Heat released = -1676 x 0.15

Heat released = 254 kJ

so the answer is option C

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