Đáp án:

 Câu 1`:m=6,4(g)`

Câu 2

`=>V_{CH_4}=\frac{672}{205}(l)`

Câu 3

`V_{CH_4}=4,48(l)`

Giải thích các bước giải:

 Câu 1

`Al_4C_3+12HCl->4AlCl_3+3CH_4`

`n_(CH_4)=\frac{2,24}{22,4}=0,1(mol)`

Theo phương trình

`n_{Al_4C_3}=1/3 n_{CH_4}=\frac{1}{30}(mol)`

Mà `H%=75%`

`=>n_{Al_4C_3}=\frac{\frac{1}{30}}{75%}=\frac{2}{45} (mol)`

`=>m=m_{Al_4C_3}=\frac{2}{45}.144=6,4(g)`

Câu 2

`n_{CH_3COONa}=\frac{15}{82}(mol)`

$CH_3COONa+NaOH\xrightarrow[CaO]{t^o}CH_4+Na_2CO_3$

Theo phương trình

`n_{CH_4}=n_{CH_3COONa}=\frac{15}{82}(mol)`

Mà `H%=80%`

`=>n_{CH_4}=\frac{15}{82}.80%=\frac{6}{41}(mol)`

`=>V_{CH_4}=\frac{6}{41}.22,4=\frac{672}{205}(l)`

Câu 3

`n_{Al_4C_3}=\frac{12}{144}=\frac{1}{12}(mol)`

`Al_4C_3+12HCl->4AlCl_3+3CH_4`

Theo phương trình

`n_{CH_4}=3n_{Al_4C_3}=0,25(mol)`

Mà `H=80%`

`n_{CH_4}=0,25.80%=0,2(mol)`

`=>V_{CH_4}=0,2.22,4=4,48(l)`

Em tham khảo nha :

\(\begin{array}{l}
6)\\
A{l_4}{C_3} + 12HCl \to 4AlC{l_3} + 3C{H_4}\\
{n_{C{H_4}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{A{l_4}{C_3}}} = \dfrac{{{n_{C{H_4}}}}}{3} = \dfrac{{0,1}}{3} = \frac{1}{{30}}mol\\
{m_{A{l_4}{C_3}}} = \dfrac{1}{{30}} \times 144 = 4,8g\\
H = 75\%  \Rightarrow {m_{A{l_4}{C_3}}}\text{ thực tế} = \dfrac{{4,8 \times 100}}{{75}} = 6,4g\\
7)\\
C{H_3}COONa + NaOH \to C{H_4} + N{a_2}C{O_3}\\
{n_{C{H_3}COONa}} = \dfrac{{15}}{{82}} \approx 0,18mol\\
{n_{C{H_4}}} = {n_{C{H_3}COONa}} = 0,18mol\\
{V_{C{H_4}}} = 0,18 \times 22,4 = 4,032l\\
H = 80\%  \Rightarrow {V_{C{H_4}}}\text{ thực tế} = 4,032 \times 80\%  = 3,2256l\\
8)\\
A{l_4}{C_3} + 12HCl \to 4AlC{l_3} + 3C{H_4}\\
{n_{A{l_4}{C_3}}} = \dfrac{m}{M} = \dfrac{{12}}{{144}} = \dfrac{1}{{12}}mol\\
{n_{C{H_4}}} = 3{n_{A{l_4}{C_3}}} = 0,25mol\\
{V_{C{H_4}}} = 0,25 \times 22,4 = 5,6l\\
H = 80\%  \Rightarrow {V_{C{H_4}}}\text{ thực tế}= 5,6 \times 80\%  = 4,48l
\end{array}\)