Given the following equation 2 CH_3OH(l) + 3 O_(g) rightarrow 2 CO_2(g) + 4H_2O(

Question

Given the following equation 2 CH_3OH(l) + 3 O_(g) rightarrow 2 CO_2(g) + 4H_2O(g) Delta H degree = 1276.97 kJ a. Identify the limiting reactant, when 0.550 L of liquid methanol (CH_3OH) is mixed with 600 g of oxygen. The density of methanol is 0.791 g/mL; the molar masses of methanol and oxygen are 32.04 g/mol and 32.00 g/mol, respectively. b. Based on the above equation, how many kilojoules of heat will be evolved when 2.50 moles of methanol is used during the reaction at standard state conditions? c. The molar enthalpies of formation of CO_2 (g) and H_2O(g) are -393.5 kJ/mol and -241.8, kJ/mol, respectively. Based on the above equation, calculation the molar enthalpy of formation of liquid methanol.

Explanation / Answer

Ans 6 a).

2 moles of methanol needs 3 moles of oxygen to react completely.

Density = mass/ volume

mass = D x V

Mass of methanol = 0.791 x 550 = 435.05 grams

number of moles = 435.05 / 32.04 = 13.58 moles

Moles of oxygen = 600/32 = 18.75 moles

13.58 moles of methanol will require (13.58 x 3) / 2 = 20.37 moles of oxygen

so oxygen is the limiting reagent here .

B) 2 moles of methanol has the enthalpy change = -1276.97 KJ

so 2.5 moles will have (2.5 x -1276.97) / 2 = -1596.21 KJ

C) enthalpy of formation of O2 = 0

Lets assume the molar enthalpy of CH3OH be x

so now

2x + 0 + 2(-393.5) + 4( -241.8) = -1276.97

2x + (-787) + (-967.2) = -1276.97

2x = 477.23

x = 238.62 KJ

So the molar enthalpy of formation of methanol = 238.62 KJ/mol

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