Đáp án:

Giải thích các bước giải:

`2Al+6HCl→2AlCl_3+3H_2↑`

`a.`

`n_{Al}=\frac{4,05}{27}=0,15(mol)`

`n_{AlCl_3}=n_{Al}=0,15(mol)`

`m_{AlCl_3}=0,15.133,5=20,025(g)`

`n_{H_2}=3/2n_{Al}=0,225(mol)`

`V_{H_2}=0,225.22,4=5,04(l)`

`b.`

`n_{PbO}=\frac{66,9}{223}=0,3(mol)`

`PbO+H_2→Pb+H_2O`

`\frac{0,3}1>\frac{0,225}1⇒PbO` dư

`n_{PbOdư}=0,3-0,225=0,075(mol)`

`n_{Pb}=n_{H_2}=0,225(mol)`

`m_{PbOdư}=0,075.223=16,725(g)`

`m_{Pb}=0,225.207=46,575(g)`

`#Devil`

Đáp án:

\( {m_{AlC{l_3}}} = 20,025{\text{ mol}}\)

\({V_{{H_2}}} = 5,04{\text{ lít}}\)

\({m_{rắn}} = 63,3{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)

Ta có:

\({n_{Al}} = \frac{{4,05}}{{27}} = 0,15{\text{ mol}} = {n_{AlC{l_3}}}\)

\({n_{{H_2}}} = \frac{3}{2}{n_{Al}} = \frac{3}{2}.0,15 = 0,225{\text{ mol}}\)

\( \to {m_{AlC{l_3}}} = 0,15.(27 + 35,5.3) = 20,025{\text{ mol}}\)

\({V_{{H_2}}} = 0,225.22,4 = 5,04{\text{ lít}}\)

Cho lượng \(H_2\) khử \(PbO\)

\(PbO + {H_2}\xrightarrow{{{t^o}}}Pb + {H_2}O\)

Ta có:

\({n_{PbO}} = \frac{{66,9}}{{207 + 16}} = 0,3{\text{ mol > }}{{\text{n}}_{{H_2}}}\)

Vậy \(PbO\) dư

Sau phản ứng rắn gồm

\({n_{Pb}} = {n_{{H_2}}} = 0,225{\text{ mol;}}{{\text{n}}_{PbO{\text{ dư}}}} = 0,3 - 0,225 = 0,075{\text{ mol}}\)

\({m_{rắn}} = {m_{Pb}} + {m_{PbO}} = 0,225.207 + 0,075.223 = 63,3{\text{ gam}}\)