Răspuns:
1.se aplica relatia
pV=niuRT---> V=niuRT/P=100X0,082X(273+27)/5=492L
2
niuCl2= V/Vm= 448/22,4= 20mol
H2 + Cl2---> 2HCl--------> niuH2=niuCl2=20mol
V= niuRT/P=20x0,082x273/4=111,9 l
3.
d=M,O2/M,H2= 32/2=16
4.
V=896mcub=896000 l
niu=V/Vm= 896000/22,4=32000mol
5
.niu=5kmol=5000mol
V= 5000x0,082x273/5=22386 l
6.
niuO2=22,4/22,4=>1mol
O2+ C---> CO2-------------------> niu,CO2=niuO2=1mol
V= 1X0,082X300/1=24,6 L
Explicație: