дано
m(Al2O3) = 204 g
m(Al)-?
4Al+3O2-->2Al2O3
M(Al2O3) =102 g/mol
n(Al2O3) = m/M = 204/ 102 = 2 mol
4n(Al) = 2n(AL2O3)
n(Al) = 4 mol
M(Al) = 27 g/mol
m(Al) = n*M = 4*27 = 108 g
ответ 108 г
дано
m(Al2O3) = 204 g
m(Al)-?
4Al+3O2-->2Al2O3
M(Al2O3) =102 g/mol
n(Al2O3) = m/M = 204/ 102 = 2 mol
4n(Al) = 2n(AL2O3)
n(Al) = 4 mol
M(Al) = 27 g/mol
m(Al) = n*M = 4*27 = 108 g
ответ 108 г