Make a summary of this text: 1) CsHg t 50,→3CO, + 4H0
2) n(CsHe) = m(C3He)/M[C3He]
m(CsH) = 13 kg = 13000 g
MIC,Ha]= 3xM[C] + 8M[H] = 3x12 + 8x1 = 44
g.mol-1
n(CHs) = 13000/44 = 295,45 mol
3) D'après l'équation, n(02) =5x n(CạHe)
Make a summary of this text: v(O) = n(O2) x Vm = 886,36 x /24 = 21272,72
5) m(CO-) = n(CO,) x M[CO,]
n(CO,) = 886,36
M[CO,] = 12 + 2x16 = 44 g.mol-
→ m(CO,) = 886,36 x 44 = 39000 g = 39 kg
m(H,0) = n(H,0) x M[H,0]
n(H,0) = 4 x n(CaHs) = 3 x 295,45 = 886,36
mol
M[H,O]= 2x1 + 16 = 18 g.mol-1
m(H,0) = 886,36 x 18 = 15954 g = environ
16 kg.
→ n(O>) = 5 x 295,45 = 1477,27 mol
Il se forme n(CO) =3 x n(CsHa) =3 x 295,45
= 886,36 mol
4) Vm = 24 L.mo|-1.
