Дано
m(Al(OH)3)=160 g
m(HCL)=200 g
---------------------
m(AlCL3)-?
M(AL(OH)3)=78 g/mol
M(HCL)=36.5 g/mol
n(Al(OH)3)=m/M=160/78=2.05 mol
n(HCL)=m/M=200/36.5=5.48 mol
n(Al(OH)3)M(AlCL3)=133.5 g/mol
160 X
AL(OH)3+3HCL-->ALCL3+3H2O
78 133.5
X=160*133.5/78=273.8 g
ответ 273.8 г