P = 700 torr , 1 atm = 760 torr
=> P = 0,92 atm
T = 273+30 = 303 K
50 g n moli
2Al + 6HCl --> 2AlCl3 + 3H2
2x27 3
=> n = 50x3/2x27 = 2,778 moli H2 obtinuti in conditii normale dar volumul in conditiile dater in problema este...
PV = nRT => V = nRT/P
= 2,778x0,082x303/0,92
= 75 L H2