Bài nào em nhỉ?

\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right);n_{Ba\left(OH\right)_2}=0,3\left(mol\right)\)

Đặt \(\left\{{}\begin{matrix}n_{Fe_3O_4}=a\left(mol\right)\\n_{Cu\left(p\text{ư}\right)}=b\left(mol\right)\end{matrix}\right.\)

Ta có sơ đồ phản ứng: \(X\left\{{}\begin{matrix}Fe_3O_4:a\\Cu:b+0,05\end{matrix}\right.+HCl\rightarrow\left\{{}\begin{matrix}FeCl_2:3a\\CuCl_2:b\\HCl\end{matrix}\right.+Cu:0,05+H_2O\)

\(\left\{{}\begin{matrix}FeCl_2:3a\\CuCl_2:b\\HCl\end{matrix}\right.\xrightarrow[]{+0,3\left(mol\right)Ba\left(OH\right)_2}18,4\left(g\right)\downarrow\left\{{}\begin{matrix}Fe\left(OH\right)_2:3a\\Cu\left(OH\right)_2:b\end{matrix}\right.+BaCl_2:0,3+H_2O\)

Quá trình oxi hóa - khử:

\(3Fe^{+\dfrac{8}{3}}+2e\rightarrow3Fe^{+2}\)

3a-------->2a

\(Cu^0\rightarrow Cu^{+2}+2e\)

b---------------->2b

BTe: 2a = 2b => a = b

\(\Rightarrow3a.90+98a=18,4\Leftrightarrow a=0,05\left(mol\right)\)

\(\Rightarrow m=0,05.232+\left(0,05+0,05\right).64=18\left(g\right)\)

BTNT Cl: \(n_{HCl}=2n_{BaCl_2}=0,6\left(mol\right)\)

\(\Rightarrow V=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)

\(n_{C_3H_8}=\dfrac{8,8}{44}=0,2\left(mol\right)\Rightarrow n_{C_3H_8\left(p\text{ư}\right)}=0,2.60\%=0,12\left(mol\right)\\ n_{C_3H_6\left(d\text{ư}\right)}=0,2-0,12=0,08\left(mol\right)\)

Quy đổi A thành \(\left\{{}\begin{matrix}C_aH_{2a+2}\\C_bH_{2b}\\C_3H_{8\left(d\text{ư}\right)}\end{matrix}\right.\left(a+b=4;a,b\in N;a\ge1;b\ge2\right)\)

PTHH: \(C_3H_8\xrightarrow[]{cracking}C_aH_{2a+2}+C_bH_{2b}\)

             0,12----------->0,12-------->0,12

BTKL: \(m_A=m_{C_3H_8}=8,8\left(g\right)\)

\(\Rightarrow\overline{M}_A=\dfrac{8,8}{0,12+0,12+0,08}=27,5\left(g/mol\right)\)

a) \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\) (cái này bạn viết ngược)

b) `Fe + 2HCl -> FeCl_2 + H_2`

c) `Na_2O + H_2O -> 2NaOH`

d) `SO_3 + H_2O -> H_2SO_4`

e) `CaO + 2HNO_3 -> Ca(NO_3)_2 + H_2O`

f) \(2Fe\left(OH\right)_3\xrightarrow[]{t^o}Fe_2O_3+3H_2O\)

g) `Na_2CO_3 + Ca(OH)_2 -> CaCO_3 + 2NaOH`

h) `CaO + CO_2 -> CaCO_3`

Cái đầu bị ngước rồi thì phải

\(CuO+H_2-^{t^o}>Cu+H_2O\\ Fe+2HCl->FeCl_2+H_2\\ Na_2O+H_2O->2NaOH\\ SO_3+H_2O->H_2SO_4\)

PTHH: \(C_4H_{10}\xrightarrow[]{cracking}C_aH_{2a+2}+C_bH_{2b}\left(a+b=4;a,b\in N;a\ge1;b\ge2\right)\)

Cứ mỗi 1 mol C₄H₁₀ pư sẽ tạo ra 2 mol sản phẩm => tăng 1 mol => mol tăng cũng là mol C₄H₁₀ pư

Ta có: 

\(\Delta n_{\uparrow}=n_{C_4H_{10}\left(p\text{ư}\right)}=540-300=240\left(mol\right)\\ \Rightarrow H=\dfrac{240}{300}.100\%=80\%\)

\(n_{NaOH}=0,1.3=0,3\left(mol\right)\)

PTHH: \(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\)

                 0,3---------------------------------------->0,45

`=> V = 0,45.22,4 = 10,08 (l)`

Bảo toàn Cacbon: \(n_{CaCO_3}=0,2\left(mol\right)=n_{CO_2}\) \(\Rightarrow V_{CO_2}=0,2\cdot22,4=4,48\left(l\right)\)

\(M_Y=14,5.2=29\left(g/mol\right)\)

Đặt \(n_X=1\left(mol\right)\)

PTHH: \(X\xrightarrow[t^o,xt]{}anken+H_2\)

           1------------>1-------->1

\(\Rightarrow m_X=m_Y=2.29=58\left(g/mol\right)\)

Đặt CTPT của X là C_nH_2n+2 (n ∈ N*)

`=> 14n + 2 = 58 <=> n = 4 (t//m)`

Vậy X là C₄H₁₀ 

Câu 1: C

Câu 2: D

Câu 3: A

Câu 4: C

Câu 5: C

Câu 6: D
Câu 7: A

Câu 8: B
Câu 9: C

Câu 10: C

\(M_Y=18.2=36\left(g/mol\right)\)

Đặt \(n_X=1\left(mol\right)\)

PTHH: \(X\xrightarrow[]{crackinh}ankan+anken\)

            1---------->1---------->1

\(\Rightarrow m_X=m_Y=2.36=72\left(g\right)\\ \Rightarrow M_X=\dfrac{72}{1}=72\left(g/mol\right)\)

Đặt CTPT của X là C_nH_2n+2 (n ∈ N*)

`=> 14n + 2 = 72 => n = 5 (t//m)`

Vậy X là C₅H₁₂