Question

L-malic acid is a compound whose salt (i.e. L-malate) is an intermediate in the tricarboxylic acid (TCA) cycle. Given the reaction and information shown below, how much thermal energy would be required to run the reaction with 4.0 g of carbon dioxide and 1.5 g of water?

4 CO2(g) + 3 H2O(l) → L-malic acid (C4H6O5(s)) + 3 O2(g) ΔHrxn = +320.1 kcal

Answer 1

Given reaction is

4 CO2(g) + 3 H2O(l) → L-malic acid (C4H6O5(s)) + 3 O2(g) ΔHrxn = +320.1 kcal

4 mole 3 mole 1 mole 3 mole   

Given

Mass of CO2 = 4 g

Mass of water = 1.5 g

Molar mass of CO2 = 44 g/mol

Molar mass of water = 18 g/mol

No. of moles of CO2 = Mass / Molar mass = 4 g / 44 g/mol = 0.091 moles

No. of moles of water = 1.5 g / 18 g/mol = 0.083 moles

4 moles of CO2 require 3 mole of water according to given reaction

so 0.091 moles of CO2 will require 0.0682 moles of H2O we have excess of H2O

so CO2 is the limiting reagent.

According to given data 4 moles of CO2 require 320.1 kcal of energy to react with 3 moles of H2O and produce L-malic acid

so 0.091 moles of CO2 will require (320.1 kcal * 0.091/4) = 7.28 kcal

Energy required = 7.28 kcal