Question

Ma puteti ajuta,va rog la problema 11?

Answer 1

%C = mC/m antracit * 100
m C = 5*95/100 = 4,75 kg C
C + O2 = CO2
1 kg C.........35,35 MJ
4,75 kg C................x
x = 167,9125 MJ
M CO2 = 44 g/mol
12 g C...........44 g CO2
4,75 kg C............y
y = 17,416 kg CO2
n = m/M = 17,416/44 = 0,395 kmoli CO2
1 mol CO2..........22,4 l CO2
0,395 kmoli CO2........z
z = 8,848 m3 CO2