дано
m(AlCL3) = 100 g
m(Al)-?
AlCL3+3K-->3KCL+Al
M(ALCL3) =133.5 g/mol
n(ALCL3) = m/M = 100 / 133.5 = 0.75 mol
n(ALCL3) = n(AL) = 0.75 mol
M(Al) = 27 g/mol
m(Al) = n*M= 0.75 * 27 = 20.25 g
ответ 20.25 г
дано
m(AlCL3) = 100 g
m(Al)-?
AlCL3+3K-->3KCL+Al
M(ALCL3) =133.5 g/mol
n(ALCL3) = m/M = 100 / 133.5 = 0.75 mol
n(ALCL3) = n(AL) = 0.75 mol
M(Al) = 27 g/mol
m(Al) = n*M= 0.75 * 27 = 20.25 g
ответ 20.25 г