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.

\(39,6g?\\ CTPT\left(A\right):C_xH_y\\ n_C=n_{CO_2}=\dfrac{39,6}{44}=0,9mol\\ m_C=0,9.12=10,8g\\ m_H=11,7-10,8=0,9g\\ M_A=2,689.29=77,981g/mol\)
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 

\(2CuO+C\xrightarrow[]{t^0}2Cu+CO_2\\ 2Fe_2O_3+3C\xrightarrow[]{t^0}4Fe+3CO_2\\ 2Al_2O_3+3C\xrightarrow[]{t^0}4Al+3CO_2\\ CO_2+CaO\rightarrow CaCO_3\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ Ca\left(OH\right)_2+2Al+2H_2O\rightarrow Ca\left(AlO_2\right)_2+3H_2\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ Fe+HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\\ Ca\left(AlO_2\right)_2+2HCl+2H_2O\rightarrow2Al\left(OH\right)_3+CaCl_2\\ Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O\\ 2Cu+O_2\xrightarrow[t^0]{}2CuO\)

\(a)n_{HCl}=\dfrac{47,45}{36,5}=1,3mol\\ n_{CuO}:n_{Fe_2O_3}=1:1\\ \Rightarrow n_{CuO}=n_{Fe_2O_3}\\ 80n_{CuO}+160n_{Fe_2O_3}=24\\ \Rightarrow80n_{CuO}+160n_{CuO}=24\\ \Rightarrow n_{CuO}=n_{Fe_2O_3}=0,1mol\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ 0,1.......0,2.........0,1..........0,1\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ 0,1..........0,6.........0,2..........0,3\\ n_{HCl.pứ}=0,2+0,6=0,8< n_{HCl}\left(1,3\right)\)
Vậy hh X tan hết
\(b)m_{CuCl_2}=0,1.135=13,5g\\ m_{FeCl_3}=0,2.162,5=32,5g\\ c)n_{HCl.dư}=1,3-0,8=0,5mol\\ ZnO+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{0,5}{2}=0,25mol\\ m_{ZnO}=0,25.81=20,25g\)

\(a)n_C=\dfrac{30}{12}=2,5mol\\ n_P=\dfrac{37,2}{31}=1,2mol\\ n_{O_2}=\dfrac{80}{22,4}=\dfrac{25}{7}mol\\ C+O_2\xrightarrow[]{t^0}CO_2\\ 4P+5O_2\xrightarrow[]{t^0}2P_2O_5\\ n_{O_2.cần,.dùng}=2,5+1,2\cdot\dfrac{5}{4}=4mol< n_{O_2}\left(\dfrac{25}{7}\right)\)
Vậy hh Y không cháy hết
\(b)2H_2O\xrightarrow[điện]{phân}2H_2+O_2\\ n_{H_2O}=4.2=8mol\\ m_{H_2O}=8.16=128g\)