a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

⇒ m_FeO = 12,6 - 5,4 = 7,2 (g)

c, Phần này đề cho dd NaOH dư hay vừa đủ bạn nhỉ?

d, Cho hh vào dd H₂SO₄ đặc nguội thì có khí thoát ra.

PT: \(2FeO+4H_2SO_{4\left(đ\right)}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=\dfrac{1}{2}n_{FeO}=0,05\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ Ta\ có :\\ m_O = m_B - m_{hh} = 5,4 - 4,44 = 0,96(mol)\\ n_O = \dfrac{0,96}{32} = 0,03(mol)\\ \Rightarrow n_{Al_2O_3}= \dfrac{1}{3}n_O = 0,01(mol)\\ \Rightarrow n_{Al} = 2n_{Al_2O_3} = 0,02(mol)\\ m_{Al} = 0,02.54 = 1,08(gam)\\ m_{Fe} = 4,44 - 1,08 = 3,36(gam)\)

Ta có n_AlCl3 = \(\dfrac{8,1}{133,5}\) = 0,06 ( mol )

2Al + 3Cl₂ → 2AlCl₃

0,06→0,09→0,06

=> m_Al = 0,06 . 27 = 1,62 ( gam )

Cho mình xin lỗi bài ở dưới bị sai mình xin giải lại

Vì Al tác dụng với HCl mà thu được chất rắn tăng 8,1 gam (AlCl₃)

Theo định luật bảo toàn khối lượng => 8,1 là khối lượng của Cl₂

=> n_Cl2 = \(\dfrac{8,1}{35,5}\) = \(\dfrac{81}{355}\) ( mol )

2Al + 3Cl₂ → 2AlCl₃

\(\dfrac{54}{355}\)\(\leftarrow\) \(\dfrac{81}{355}\)

=> m_Al = \(\dfrac{54}{355}\) . 27 \(\approx\) 4,11 ( gam )

a)

$V_{CH_4} = V_{khí\ thoát\ ra}  = 2,24(lít)$
$\%V_{CH_4} = \dfrac{2,24}{8,96}.100\% = 25\%$

$\%V_{C_2H_4} = 100\% -25\% = 75\%$

b)

$n_{Br_2} = n_{C_2H_4} = \dfrac{8,96.75\%}{22,4} = 0,3(mol)$
$C_{M_{Br_2}} = \dfrac{0,3}{0,2} = 1,5M$
$m_{tăng} = m_{C_2H_4} = 0,3.28 = 8,4(gam)$

\(a,n_{hhkhí\left(C_2H_4,C_2H_2\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{80}{160}=0,5\left(mol\right)\\ Gọi\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ Hệ.pt\left\{{}\begin{matrix}a+b=0,3\\a+2b=0,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\\ \%V_{C_2H_4}=\dfrac{0,1}{0,3}=33,33\%\\ \%V_{C_2H_2}=100\%-33,335=66,67\%\)

\(b,PTHH:\\ C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:0,1\rightarrow0,3\rightarrow0,2\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:0,2\rightarrow0,25\rightarrow0,4\\ n_{CO_2}=0,2+0,4=0,6\left(mol\right)\\ PTHH:Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ Mol:0,6\rightarrow0,6\rightarrow0,6\\ m_{CaCO_3}=0,6.100=60\left(g\right)\)

\(a,Gọi\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{C_2H_2}=c\left(mol\right)\end{matrix}\right.\\ n_{hhkhí}=0,4\left(mol\right)\\ n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:a\rightarrow2a\rightarrow a\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:b\rightarrow3b\rightarrow2b\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:c\rightarrow2,5c\rightarrow2c\\ Hệ.pt\left\{{}\begin{matrix}a+b+c=0,4\\b+2c=0,4\\a+2b+2c=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\)

\(\%V_{CH_4}=\%V_{C_2H_2}=\dfrac{0,1}{0,4}=25\%\\ \%V_{C_2H_4}=\dfrac{0,2}{0,4}=50\%\)

\(m_{CH_4}=0,1.16=1,6\left(g\right)\\ m_{C_2H_4}=28.0,2=5,6\left(g\right)\\ m_{C_2H_2}=0,1.26=2,6\left(g\right)\\ \%m_{CH_4}=\dfrac{1,6}{1,6+5,6+2,6}=16,32\%\\ \%m_{C_2H_4}=\dfrac{5,6}{1,6+5,6+2,6}=57,14\%\\ \%m_{C_2H_2}=100\%-16,32\%-57,14\%=26,54\%\)

\(b,PTHH:C_2H_5OH\rightarrow C_2H_4+H_2O\\ Mol:0,2\leftarrow0,2\\ m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)

Dài quá!!!