дано
n(al) = 4 mol
m(F2)-?
3F2+2Al-->2AlF3
n(F) = 3*4 / 2 = 6 mol
M(F2) = 19*2 = 38 g/mol
m(F2) = n*M = 6 * 38 = 228 g
ответ 228 г
