x y 360
H₂ + 1/2O₂ ⇒ H₂O
2 32/2 18
x=[tex] \frac{360*2}{18}[/tex]= 40 g H₂
y= [tex] \frac{360*16}{18} [/tex]= 320 g O₂
Aflii numarul de moli pentru fiecare element: n=[tex] \frac{m}{M} [/tex]
n indice H₂=[tex] \frac{40}{2} [/tex]= 20 moli H₂
n indice O₂=[tex] \frac{320}{32} [/tex]= 10 moli O₂
Ecuatia de stare a gazelor: p*V=n*R*T
p= 2atm
T=127°C+273= 400K (Kelvin)
R= 0,082 L*atm * mol⁻¹ * K⁻¹
V=[tex] \frac{n*R*T}{p} [/tex]
VO₂=[tex] \frac{ n_{O2}* R*T }{p}= \frac{10*0,082*400}{2}=164L [/tex] O₂
VH₂=[tex] \frac{ n_{H2} *R*T}{p}= \frac{20*0,082*400}{2}= 328 L H_{2} [/tex]
