\(2Al+3Cl_2\underrightarrow{^{to}}2AlCl_3\)
\(n_{Al}=\frac{1,35}{27}=0,05\left(mol\right)\)
\(\Rightarrow n_{Cl2}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,05=0,075\left(mol\right)\)
\(\Rightarrow V_{Cl2}=0,075.22,4=1,68\left(l\right)\)
\(n_{AlCl3}=n_{Al}=0,05\left(mol\right)\)
\(\Rightarrow m_{AlCl3}=0,05.133,5=6,675\left(g\right)\)
a) PTHH:
2Al + 3Cl2 -> 2AlCl3
_Biết làm mỗi câu này thôi, thông cảm!_
2Al+3Cl2-->2AlCl3
0,05--0,075--0,05 mol
nAl=1,35\27=0,05 mol
=>VCl2=0,075.22,4=1,68l
=>mAlCL3=0,05.133,5 =6,675 g