2Al+ 3H2SO4 -> Al2(SO4)3 + 3H2
x g Al....................2.8 g H2SO4
2*27 g Al................3*98 g H2SO4 ⇒
x=[tex]x= \frac{27*2*2.8}{3*98}= 0.5 g [/tex] moli Al: 0.01 moli Al folosit
[tex] mAl_{exces}=5.4-0.5=4.9 g [/tex]
98*3 g H2SO4 ......................3*2 g H2
2.8 g H2SO4..........................y g H2
[tex]y= \frac{2.8*6}{3*98}=0.05 g H2 iese din reactie
[/tex]
moli H2: 0.05/2=0.025 moli
V_{H2}= 22.4*0.025=0.56 l H2 in conditii normale
