Đáp án:

Câu 6: mAl=2,7g

V$Cl_{2}$=3,36l

Câu 7: 1,12l

Câu 8: 78%

Giải thích các bước giải:

Câu 6:

$nAlCl_{3}$=$\frac{13,35}{133,5}$=0,1

Ta có \(2Al + 3C{l_2} \to 2AlC{l_3}\)

nAl=\(nAlCl_{3}\)=0,1

\(\to\)mAl=0,1.27=2,7g

$nCl_{2}$=nAl.$\frac{3}{2}$=0,1.$\frac{3}{2}$=0,15

->V$Cl_{2}$=0,15.22,4=3,36l

Câu 7:

n$KMnO_{4}$=$\frac{15,8}{158}$=0,1

Ta có: \(2KMnO_{4} \to K_{2}MnO_{4}+MnO_{2}+O_{2}\)

$nO_{2}$=\(nKMnO4\).0,5=0,1.0,5=0,05

\(\to\)V$O_{2}$=0,05.22,4=1,12l

Câu 8:

Ta có \(H_{2}+Cl_{2}\to2HCl\)

nHCl thu được=$\frac{10,92}{22,4}$=0,4875

\(\to\)$nCl_{2}$ phản ứng=$\frac{nHCl}{2}$=0,24375

\(\to\)$VCl_{2}$ phản ứng=0,24375.22,4=5,46l

\(\to\)H=$\frac{5,46}{7}$=78%

`1.`

`n_{AlCl_3}={13,35}/{133,5}=0,1(mol)`

`2Al+3Cl_2`  $\xrightarrow{t^o}$  `2AlCl_3`

`n_{Al}=n_{AlCl_3}=0,1(mol)`

`m_{Al}=0,1.27=2,7(g)`

`n_{Cl_2}=3/2n_{Al}=0,15(mol)`

`V_{Cl_2}=0,15.22,4=3,36(l)`

`2.`

`n_{KMnO_4}={15,8}/{158}=0,1(mol)`

`2KMnO_4+16HCl->2KCl+2MnCl_2+5Cl_2+8H_2O`

`n_{Cl_2}=5/2n_{KMnO_4}=0,25(mol)`

`V_{Cl_2}=0,25.22,4=5,6(l)`

`3.`

`H_2+Cl_2`  $\xrightarrow{t^o}$  `2HCl`

`n_{Cl_2(t t)}=1/2n_{HCl}`

`->V_{Cl_2(t t)}=1/2V_{HCl}=5,46(l)`

`->H={5,46}/7.100\%=78\%`