What is the molar heat of combustion of methanol, CH_3OH, if combustion of 1.00

Question

What is the molar heat of combustion of methanol, CH_3OH, if combustion of 1.00 g of methanol causes a temperature rise of 3.68 oC in a bomb calorimeter that has a heat capacity of 6.43 kJ/oC?(Formula weight 55.9 kJ/mol 923 kJ/mol 757 kJ/mol 18.3 kJ/mol 368 kJ/mol From a consideration of the reaction 2NO(g) + O2(g) rightarrow 2NO_2(g) H = 114.1 kJ if2.00 times 10^2 g of NO2 were produced, then the amount of heat released should be 114 kJ. 124 kJ. 248 kJ. 314 kJ. 496 kJ. The equation for the standard enthalpy of formation of potassium bromate, KBrO_3, corresponds to which reaction? K(s) + 1/2 Br_2(g) + 3/2 O_2(g) rightarrow KBrO_3(s)

Explanation / Answer

34) mol of methanol = 1 gm / 32 gm/mol = 0.03125 mol
   energy released from combustion = 3.68o C * 6.43 kJ/oC = 23.6624 kJ
   molar heat of combustion of methanol = 23.6624 kJ / 0.03125 mol = 757.19 kJ/mol
   Option C is the answer

35) According to the stoichiometry,
(2 *102 gm / 46.0055 gm) * (1 mol of NO2 / 2 mol of NO2) * (114.1 kJ) = 248.0138 kJ
So, option C is the answer

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