Balance the chemical equation below: CO2 + H2O + 2678 kJ-> C6H1206 + O2 a. How many moles of H20 are involved in this reaction? b. Is this reaction Exothermic or Endothermic? c. How much energy will be involved in this reaction when 1 mole of CO2 is used up?

a) 6 mol H2O

b) this reaction is endothermic

c) when 1 mol of CO2 is used, in the reaction they occur 0.5025 KJ

Explanation:

balanced eq:

6CO2 + 6H2O + 2678 KJ ↔ C6H12O6 + 6O2

6 - C - 6

18 - O - 18

12 - H - 12

a) mol H2O = 6 mol ... from balanced equation.

b) ΔE = 2678 KJ ... this reaction absorbs heat (ΔE is positive)

c) 1 gramo C6H12O6 ≅ 4 cal

Mw C6H12O6 = 180.156 g/mol

⇒ 1mol CO2 * (mol C6H12O6 / 6mol CO2) = 0.166 mol C6H12O6

⇒ 0.166mol C6H12O6 * (180.156 g C6H12O6 / mol) = 30.026g C6H12O6

⇒30.026 gC6H12O6 * (4 cal / gC6H12O6) * (Kcal / 1000 cal) * (4184 J / Kcal) * (KJ / 1000 J) = 0.5025 KJ C6H12O6.