Tolong dong yg bisa No 9 sama 10 Terimakasih :)

9) Dik : nEtanol = 92/46 = 2 mol
[tex]2C + 3H_{2} + \frac{1}{2}O_{2} ===\ \textgreater \ C_{2}H_{5}OH [/tex] ΔHf = - 278 kJ/Mol
[tex]C + O_{2} ===\ \textgreater \ CO_{2}[/tex] ΔHf = - 394 kJ/Mol
[tex]H_{2} + \frac{1}{2}O_{2} ===\ \textgreater \ H_{2}O [/tex] ΔHf = -286 kJ/Mol
Dit : ΔHc [tex]C_{2}H_{5}OH + 3O_{2} ==\ \textgreater \ 2CO_{2} + 3H_{2}O [/tex]
Jawab :
[tex]C_{2}H_{5}OH ==\ \textgreater \ 2C + 3H_{2} + \frac{1}{2} O_{2} [/tex] ΔHf = + 278 kJ/Mol
[tex]2C + 2O_{2} ==\ \textgreater \ 2CO_{2}[/tex] ΔHf = - 788 kJ/Mol
[tex]3H_{2} + \frac{3}{2}O_{2} ==\ \textgreater \ 3H_{2}O [/tex] ΔHf = - 858 kJ/Mol

∴ ΔH = 278 - 788 - 858 = - 1368 kJ
   ΔHc = -1368/mol etanol
   ΔHc = - 1368/2
   ΔHc = - 684 kJ/Mol

10) Dik :
[tex]2CO + O_{2} ==\ \textgreater \ 2CO[/tex] ΔH = - 516 kJ
[tex]4MnO + O_{2} ==\ \textgreater \ 2Mn_{2}O_{3}[/tex] ΔH= - 312 kJ
Dit : ΔH [tex]Mn_{2}O_{3} + CO ==\ \textgreater \ 2MnO + CO_{2}[/tex]
Jawab :
[tex]Mn_{2}O_{3} ==\ \textgreater \ 2MnO + \frac{1}{2}O_{2} [/tex] ΔH = +156 kJ
[tex]CO + \frac{1}{2}O_{2} ==\ \textgreater \ CO [/tex] ΔH = - 258 kJ

∴ ΔH = 156 - 258
   ΔH = - 102 kJ