Question

Methanol (CH3OH) is used as a fuel in race cars.

Part A Write a balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product. Express your answer as a chemical equation. Identify all of the phases in your answer. SubmitPrevious AnswersRequest Answer Incorrect; Try Again; 5 attempts remaining Part B Calculate the standard enthalpy change for the combustion of 1 mol of liquid methanol, assuming H2O(g) as a product. Express your answer using four significant figures. ΔH∘rxn ΔHrxn∘ = nothing   kJ   SubmitRequest Answer Part C Calculate the heat produced by combustion per liter of methanol. Methanol has a density of 0.791 g/mL. Express your answer using three significant figures. q q = nothing   kJ/L   SubmitRequest Answer Part D Calculate the mass of CO2 produced per kJ of heat emitted. Express your answer using four significant figures. m m =   g/kJ

Answer 1

Answer – Part A ) We are given methanol and we need to write the balanced combustion reaction equation

2 CH₃OH (l) + 3 O₂(g) -----> 2 CO₂(g) + 4 H₂O(g)

Part B) We know standard enthalpy formula

We know,

ΔH^o_rxn = ∑ ΔH^o_f of product – ∑ ΔH^o_f of reactant

          = [ (2* ΔH^o_f CO₂(g)) + (4* ΔH^o_f H₂O(g) )] – [ 2*ΔH^o_f CH₃OH (l) + 3 *ΔH^o_f O₂(g) ]

          = (2*-393.51 + 4*-241.82) - ( 2*-239.03 + 3*0.00)

           = -1276 kJ

This is for 2 moles of CH₃OH, so for 1 mole of CH₃OH = -1276 /2 = -638.1 kJ

Part C) We are given, density of methanol = 0.791 g/mL

volume of methanol = 1 L = 1000 mL

we need to calculate the moles of methanol

mass of methanol = density x volume

                                  = 0.791 g/mL x 1000 mL

                                  = 791 g

moles of methanol = 791g / 32.04 g.mol^-1

                                   = 24.68 moles

From the above reaction

2 moles of CH₃OH = -1276.24 kJ

so, 24.68 moles of CH₃OH = ? kJ

= -1.57 x10⁴ kJ/L

Part D) From the first reaction

-1276 kJ = 2 moles of CO₂

-1 kJ = ? moles of CO₂

= 0.00157 moles of CO₂

Mass of CO₂ = 0.00157 moles x 44 g/mol

                       = 0.06895 g of CO₂