Verified answer

40 г                                                                              Х л

Mg          +        2HNO3      =       Mg(NO3)2       +       H2

n=1 моль                                                                  n = 1 моль

Mr=24 г/моль                                                           Vm = 22,4 л/моль

m = 24 г                                                                    V = 22,4 л

40 г Mg     -    Х л Н2

24 г Mg    -     22,4 л Н2

V(H2) = 40 * 22,4 / 24 = 37,3 л