Answer:
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.
