5,6 tấn = 5600 (kg)

Ta có: \(n_{Fe}=\dfrac{5600}{56}=100\left(kmol\right)\)

PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

Theo PT: \(n_{Fe_2O_3\left(LT\right)}=\dfrac{1}{2}n_{Fe}=50\left(kmol\right)\)

Mà: H = 90%

\(\Rightarrow n_{Fe_2O_3\left(TT\right)}=\dfrac{50}{90\%}=\dfrac{500}{9}\left(kmol\right)\)

\(\Rightarrow m_{Fe_2O_3\left(TT\right)}=\dfrac{500}{9}.160=\dfrac{80000}{9}\left(kg\right)\)

⇒ m _quặng = \(\dfrac{m_{Fe_2O_3\left(TT\right)}}{30\%}\approx29629,6\left(kg\right)\) ≃ 29,63 (tấn)

Ta có: m_H2O = 0,18 (g) 

\(\Rightarrow n_{H_2O}=\dfrac{0,18}{18}=0,01\left(mol\right)\Rightarrow n_H=0,01.2=0,02\left(mol\right)\)

m_CO2 = 0,44 (g)

\(\Rightarrow n_{CO_2}=\dfrac{0,44}{44}=0,01\left(mol\right)=n_C\)

Có: \(n_A=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

Gọi CTPT của A là C_xH_y.

\(x=\dfrac{0,01}{0,005}=2\)

\(y=\dfrac{0,02}{0,005}=4\)

Vậy: A là C₂H₄.

(1) \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

(2) \(FeSO_4+Ba\left(NO_3\right)_2\rightarrow BaSO_4+Fe\left(NO_3\right)_2\)

(3) \(Fe\left(NO_3\right)_2+2NaOH\rightarrow2NaNO_3+Fe\left(OH\right)_2\)

(4) \(Fe\left(OH\right)_2\xrightarrow[_{\left(kckk\right)}]{^{t^o}}FeO+H_2O\)

(5) \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

(6) \(2Fe+6H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

(6) thiếu số 2 ở Fe nhé cj

\(39,6g?\\ CTPT\left(A\right):C_xH_y\\ n_C=n_{CO_2}=\dfrac{39,6}{44}=0,9mol\\ m_C=0,9.12=10,8g\\ m_H=11,7-10,8=0,9g\\ M_A=2,689.29=77,981g/mol\)
Ta có tỉ lệ
\(\dfrac{12x}{10,8}=\dfrac{y}{0,9}=\dfrac{77,981}{11,7}\\ \Rightarrow x\approx6;y\approx6\\ \Rightarrow CTPT\left(A\right):C_6H_6\\ 2C_6H_6+15O_2\rightarrow12CO_2+6H_2O\\ n_{H_2O}=\dfrac{0,9.6}{12}=0,45mol\\ m=m_{H_2O}=0,45.18=8,1g\)

giúp mình với ạ 

giải nhanh hộ e vs ạ e cần gấp ạ 

\(HCl+NaOH\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaCl}=n_{NaOH}=1,5.0,1=0,15\left(mol\right)\\ a,m_{HCl}=0,15.36,5=5,475\left(g\right)\\ b,m_{NaCl}=58,5.0,15=8,775\left(g\right)\)

a, Ta có: \(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

BTNT H, có: n_HCl = 2n_H2 = 0,8 (mol)

BTKL, có: m_Al + m_Fe + m_HCl = m _muối + m_H2

⇒ m _muối = 11 + 0,8.36,5 - 0,4.2 = 39,4 (g)

b, \(C\%_{HCl}=\dfrac{0,8.36,5}{300}.100\%\approx9,733\%\)

c, Ta có: 27n_Al + 56n_Fe = 11 (1)

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

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=0,4\left(2\right)\)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{AlCl_3}=\dfrac{0,2.133,5}{39,4}.100\%\approx67,77\%\\\%m_{FeCl_2}\approx32,23\%\end{matrix}\right.\)

a, Gọi: n_H2 = a (mol)

BTNT H, có: n_HCl = 2n_H2 = 2a (mol)

BTKL, có: m_Na + m_Mg + m_HCl = m _muối + m_H2

⇒ 10 + 2a.36,5  = 30,7 + 2a

⇒ a = 207/710 (mol)

\(\Rightarrow C\%_{HCl}=\dfrac{2.\dfrac{207}{710}.36,5}{200}.100\%\approx10,64\%\)

b, \(V_{H_2}=\dfrac{207}{710}.24,79\approx7,23\left(l\right)\)

c, Ta có: 23n_Na + 24n_Mg = 10 (1)

PT: \(2Na+2HCl\rightarrow2NaCl+H_2\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(\left\{{}\begin{matrix}n_{NaCl}=n_{Na}\\n_{MgCl_2}=n_{Mg}\end{matrix}\right.\)

⇒ 58,5n_Na + 95n_Mg = 30,7 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na}=\dfrac{1066}{3905}\left(mol\right)\\n_{Mg}=\dfrac{1211}{7810}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{\dfrac{1066}{3905}.58,5}{10+200-\dfrac{207}{710}.2}.100\%\approx7,63\%\\C\%_{MgCl_2}=\dfrac{\dfrac{1211}{7810}.95}{10+200-\dfrac{207}{710}.2}.100\%\approx7,03\%\end{matrix}\right.\)

Số mol khá lẻ nên bạn có thể xem lại đề rồi tự làm theo cách làm như trên nhé.

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\ m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{2}\Rightarrow HCldư\\ n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,2=4,958\left(l\right)\)

\(n_{HCl}=\dfrac{10\%.109,5}{36,5}=0,3\left(mol\right);n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,3}{2}>\dfrac{0,1}{1}\Rightarrow HCldư\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.0,1=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,1=2,479\left(l\right)\\ b,ddA:HCl\left(dư\right),MgCl_2\\ m_{ddA}=2,4+109,5-0,1.2=111,7\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,1.36,5}{111,7}.100\%\approx3,268\%;C\%_{ddMgCl_2}=\dfrac{0,1.95}{111,7}.100\%\approx8,505\%\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right);n_{HCl}=\dfrac{365.10\%}{36,5}=1\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{1}{2}>\dfrac{0,1}{1}\Rightarrow HCldư\\ n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)