10 г                                                               Х л

    2Li       +       2H2O      =       2LiOH       +      H2

n = 2 моль                                                       n = 1 моль

Mr = 7 г/моль                                                Vm = 22,4 л/моль

m = 14 г                                                         V = 22,4 л

10 г Li      -      Х л Н2

14 г Li     -       22,4 л Н2

V(H2) = 10 * 22,4 / 14 = 16 л