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Jawab No.1

Mol = [tex] \frac{Massa}{Ar atau Mr} [/tex]
    
       =[tex] \frac{75g}{2.ArC+6.ArH } [/tex]
    
       =[tex] \frac{75g }{2.12+6.1} [/tex]
  
       =[tex] \frac{75g}{30} [/tex]
      
       =2,5 Mol

PV = nRT
V   = [tex] \frac{nRT}{P} [/tex]
    
     =[tex] \frac{2,5mol.0,82LatmMolK.300K}{5atm} [/tex]
  
     =12,3 Liter



Jawab No.2
Cara coba-coba:
H2SO4 + Al(OH)3 → Al2(SO4)3 + H2O
3          2                  1              6

Cara substitusi:

aH2SO4 + bAl(OH)3 → cAl2(SO4)3 + dH2O
H : 2a + 3b = 2d
S : a = 3c
O : 4a + 3b = 12c + d
Al : b = 2c
Misal: c = 1, maka: a = 3; b = 2 
2a + 3b = 2d
2x3 + 3x2 = 2xd              
D=6
Jadi, a=3;  b=2;  c=1;  d=6

1. etana/C2H6 pertama-tama cari Mr C2H6 = 30
    mol C2H6 = massa/Mr = 75/30 = 2,5 mol
    V = nRT/P = 2,5x0,0821x300/5 = 12,315 liter
2. 3 H2SO4 + 2 Al(OH)3 --> Al2(SO4)3 + 6 H2O