14,4 г                                                Х л  Ba      +      2HCl     -gt;  BaCl2   +    H2n = 1 моль                                       n = 1 мольMr=137 г/моль                               Vm=22,4 л/мольm = 137 г                                         V = 22,4 л14,4 г Ва     -      Х л Н2137 г Ва     -     22,4 л Н2V(H2) = 14,4 * 22,4 / 137 = 2,35 л