\(2CH_4\rightarrow C_2H_2+3H_2\) \(\left(ĐK:t^o\right)\)

\(2CH_4\xrightarrow[lln]{t^ocao}C_2H_2+3H_2\)

(lln: làm lạnh nhanh)

Đề chưa đủ em nhé!

PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

Gọi: \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 102x + 80y = 17,1 (1)

Ta có: \(m_{HCl}=300.7,3\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Al_2O_3}+2n_{CuO}=6x+2y=0,6\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)

PT: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

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

Gọi: \(\left\{{}\begin{matrix}n_{MgO}=x\left(mol\right)\\n_{FeO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 40x + 72y = 15,2 (1)

Ta có: n_HCl = 0,3.2 = 0,6 (mol)

Theo PT: \(n_{HCl}=2n_{MgO}+2n_{FeO}=2x+2y=0,6\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)

Ta có: \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

\(m_{H_2SO_4}=250.19,6\%=49\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\)

PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{ZnSO_4}=n_{ZnO}=0,2\left(mol\right)\)

\(\Rightarrow m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)

\(PTPU:ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

                \(0,2:0,2:0,2:0,2\left(mol\right)\)

\(n_{ZnO}=\dfrac{m}{M}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

\(m_{ZnSO_4}=n.M=0,2.161=32,2\left(g\right)\)

Ta có: n_HCl = 0,2.3 = 0,6 (mol)

PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{6}n_{HCl}=0,1\left(mol\right)\)

⇒ m_Al2O3 = 0,1.102 = 10,2 (g)

Số xấu quá em xem lại đề nha

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

a) \(n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

\(n_{HCl}=\dfrac{m}{M}=\dfrac{18,5}{36,5}=\dfrac{37}{73}\left(mol\right)\)

Lập bảng: \(\dfrac{0,4}{1}>\dfrac{\dfrac{37}{73}}{2}\)

⇒ sau pư HCl hết, Fe dư

⇒ theo \(n_{HCl}\)

Theo PTHH: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=\dfrac{\dfrac{37}{73}}{2}=\dfrac{37}{146}\left(mol\right)\)

⇒ \(n_{Fe\left(dư\right)}=0,4-\dfrac{37}{146}=\dfrac{107}{730}\left(mol\right)\)

\(\Rightarrow m_{Fe\left(dư\right)}=n.M=\dfrac{107}{730}.56=\dfrac{2996}{365}\left(g\right)\)

b) Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{\dfrac{37}{73}}{2}=\dfrac{37}{146}\left(mol\right)\)

\(\Rightarrow V_{H_2\left(đktc\right)}=n.22,4=\dfrac{37}{146}.22,4=\dfrac{2072}{365}\left(l\right)\)

c) Theo PTHH: \(n_{FeCl_2}=n_{H_2}=\dfrac{37}{146}\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=n.M=\dfrac{37}{146}.127=\dfrac{4699}{146}\left(g\right)\)

`n_Zn = m/M = 6,5/65 = 0,1 (mol) `

\(PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)

Tỉ lệ:       1     :    1        :           1      : 1

n(mol)   0,1---->0,1-------------->0,1--->0,1

\(m_{H_2SO_4}=n\cdot M=0,1\cdot\left(2+32+16\cdot4\right)=9,8\left(g\right)\)

\(V_{H_2\left(dkt\right)}=n\cdot24=0,1\cdot24=2,4\left(l\right)\)

\(m_{ZnSO_4}=n\cdot M=0,1\cdot\left(65+32+16\cdot4\right)=16,1\left(g\right)\)

\(PTPU:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

               \(0,1:0,1:0,1:0,1\left(mol\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(a,m_{H_2SO_4}=n.M=0,1.\left(2+32+16.4\right)=9,8\left(g\right)\)

\(b,V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

\(c,m_{ZnSO_4}=n.M=0,1.\left(65+32+16.4\right)=16,1\left(g\right)\)

đổi 280ml=0,28(l)

\(n_{Na\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{0,28}{22,4}=0,0125\left(mol\right)\)

\(PTHH:2Na+2H_2O->2NaOH+H_2\)

tỉ lê:       2         :   2         :    2            : 1

n(mol)   0,0125->0,0125-->0,0125

`m_Na =n*M=0,0125*23=0,2875(g)`

`m_NaOH =n*M=0,0125*(23+1+16)=0,5(g)`