Question

18. In a calorimetry experiment, 0.1277 g af Me ribbon was added to 200.0 mL 0. HCl at 24.12 °C. The water temperature increased to 27.10 °C. Calculate AH for reaction, as performed, and AH per mole of HCI. 19. A 2.50 g sample of sucrose (C2H220) was burned in excess oxygen in a car which contained 2.19 kg of water. The temperature of the water increased from 20. to 25.01 °C. Determine the molar heat of combustion of sucrose. 20. The burning of 5.08 g benzene (CH) releases enough heat to raise the temperature of 5.00 kg of water from 10.1 °C to 19.6°C. Calculate the molar heat of combustion of benzene. 21. ΔΗ -0.02 kJ Given the following equations and AH values, H;BO3(aq) - HBO2(aq) + H2O() H2B.Or(s) + H20() - 4 HBO3(aq) -11.3 kJ +17.5 kJ HzB4O-(s) - 2 B 03(s) + H2O() B2O3(s) + 3 H2O(1) calculate Ahºrn for the following reaction: 2 HBO3(aq) - 22. The following reaction is one that occurs in a blast fumace when iron is extracted from its ores: Fe2O3(s) + 3 CO(g) — 2 Fe(s) + 3 CO2(g) find AHºn for this reaction given the following information: ΔΗ 3 Fe2O3(s) + CO(g) - 2 Fe3O4(s) + CO2(g) 46.4 kJ FeO(s) + CO(g) — Fe(s) + CO2(8) 9.0 kJ Fe3O4(s) + CO(g) — 3 Fe(s) + CO2(g) 41.0 kJ

Answer 1

18) Equation: Mg(s) + 2 HCl(aq) ----> MgCl2(aq) + H2(g)

No of mole of Mg reacted = 0.1277/24 = 0.00532 mole

No of mole of HCl taken = 200/1000*0.5 = 0.1 mol

heat released(q) = m*s*DT

m = mass of reaction mixture = 200 g

s = specific heat of reaction mixture = 4.184 j/g.c

DT = 27.10-24.12 = 2.98

q = 200*4.184*2.98 = 2493.7 joule

= 2.5 kj

DHrxn per mol of HCl = -2.5/0.1 = -25 kj/mol

19)

C12H22O11(S) + 12o2(g) ----> 12CO2(g) + 11H2O(g)

heat released(q) = m*s*DT

m = mass of water = 2.5*10^3 g

s = specific heat of water = 4.184 j/g.c

DT = 25.01-20.50 = 4.51

q = 2.5*10^3*4.184*4.51 = 47174.6 joule

= 47.174 kj

no of mole of sucreose burned = 2.5/342.3 = 0.0073 mole

DHrxn = -47.174/0.0073 = -6462.2 kj/mol

20)  

heat released(q) = m*s*DT

m = mass of water = 5*10^3 g

s = specific heat of water = 4.184 j/g.c

DT = 19.6-10.1 = 9.5

q = 5*10^3*4.184*9.5 = 198740 joule

= 198.74 kj

no of mole of benzene burned = 5.08/78 = 0.065 mole

DHrxn = -198.74/0.065 = -3057.5 kj/mol