The complete combustion of 1.47g of methanol produces 29.3 kjof heat. Determine

Question

The complete combustion of 1.47g of methanol produces 29.3 kjof heat. Determine the Ho for the reaction and itssign. CH3OH(l) + 3/2 O2(g) -------->CO2(g) + 2H2O a. -938 kj b. -638 kj c. -1.35kj d.+638kj e.+938kj Alright please show step by step on how you solved this, aclear explanation will help me understand how to solve a problemlike this. If you show each step and explain how you got youranswer then I will give you full kharma points, thanks for yourtime and helping me understand ! The complete combustion of 1.47g of methanol produces 29.3 kjof heat. Determine the Ho for the reaction and itssign. CH3OH(l) + 3/2 O2(g) -------->CO2(g) + 2H2O a. -938 kj b. -638 kj c. -1.35kj d.+638kj e.+938kj Alright please show step by step on how you solved this, aclear explanation will help me understand how to solve a problemlike this. If you show each step and explain how you got youranswer then I will give you full kharma points, thanks for yourtime and helping me understand !

Explanation / Answer

The enthalpy of combustion of methanol, CH3OH, is Ho = -X kJ mol-1 CH3OH(l) + 3/2 O2(g) -------->CO2(g) + 2H2O CH3OH(l) + 3/2 O2(g) -------->CO2(g) + 2H2O Molar mass of CH3OH is = 12 + 3 * 1 + 16 + 1 = 32 g 1.47 g of methanol on complete combustion ,the Ho of the reaction is - 29.3 KJ (Since energy is released ==> -ve sign) 32 g of methanol on complete combustion ,the Ho of the reaction is X mol X = -( 29.3 * 32 ) / 1.47      = -637.82 KJ      ~ -638 KJ

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