\(n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

PT: \(Cu+2FeCl_3\rightarrow2FeCl_2+CuCl_2\)

Xét tỉ lệ: \(\dfrac{0,3}{1}=\dfrac{0,6}{2}\), ta được pư vừa đủ.

BTKL, có: m_Cu + m_FeCl3 = m _muối

⇒ m = 19,2 + 0,6.162,5 = 116,7 (g)

Sửa đề: Fe(NO₃)₂ → Fe(NO₃)₃

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(n_{Fe\left(NO_3\right)_3}=0,4.0,8=0,32\left(mol\right)\)

PT: \(Fe+2Fe\left(NO_3\right)_3\rightarrow3Fe\left(NO_3\right)_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,32}{2}\), ta được Fe dư.

Theo PT: \(n_{Fe\left(NO_3\right)_2}=\dfrac{3}{2}n_{Fe\left(NO_3\right)_3}=0,48\left(mol\right)\)

\(\Rightarrow m_{Fe\left(NO_3\right)_2}=0,48.180=86,4\left(g\right)\)

Ta có: \(n_{NO}+n_{NO_2}+n_{N_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\left(1\right)\)

Mà: m_X = 35,8 (g)

\(\Rightarrow30n_{NO}+46n_{NO_2}+28n_{N_2}=35,8\left(2\right)\)

Có: \(n_{Al}=\dfrac{32,4}{27}=1,2\left(mol\right)\)

\(n_{Cu}=\dfrac{22,4}{64}=0,35\left(mol\right)\)

BT e, có: 3n_NO + n_NO2 + 10n_N2 = 3n_Al + 2n_Cu = 4,3 (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{NO}=0,3\left(mol\right)\\n_{NO_2}=0,4\left(mol\right)\\n_{N_2}=0,3\left(mol\right)\end{matrix}\right.\)

⇒ n_HNO3 = 4n_NO + 2n_NO2 + 12n_N2 = 5,6 (mol)

Ta có: \(n_{N_2O}+n_{NO_2}+n_{N_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

BT e, có: 2n_Mg + 3n_Fe = 8n_N2O + n_NO2 + 10n_N2

⇒ 8n_N2O + n_NO2 + 10n_N2 = 1,9 (2)

Có: n_HNO3 = 10n_N2O + 2n_NO2 + 12n_N2 = 2,4.1 = 2,4 (mol) (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{N_2O}=0,1\left(mol\right)\\n_{NO_2}=0,1\left(mol\right)\\n_{N_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%V_{N_2}=\dfrac{0,1.22,4}{6,72.}100\%\approx33,33\%\)

\(n_{N_2O}=a\left(mol\right)\)

\(n_{NO_2}=b\left(mol\right)\)

\(n_{N_2}=c\left(mol\right)\)

\(\Rightarrow n_X=a+b+c=\dfrac{6.72}{22.4}=0.3\left(mol\right)\left(1\right)\)

Bảo toàn e : 

\(8a+b+10c=0.65\cdot2+0.2\cdot3=1.9\left(2\right)\)

\(n_{H^+}=10a+2b+12c=2.4\left(mol\right)\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right):a=b=c=0.1\left(mol\right)\) 

\(\%V_{N_2}=\dfrac{0.1}{0.3}\cdot100\%=33.33\%\)

\(n_{HNO_3}=1.2\left(mol\right)\)

\(n_{NO}=a\left(mol\right)\)

\(n_{NO_2}=b\left(mol\right)\)

\(n_{N_2}=c\left(mol\right)\)

\(\Rightarrow a+b+c=\dfrac{5.6}{22.4}=0.25\left(1\right)\)

Bảo toàn e : 

\(3\cdot0.2+1\cdot0.3=3a+b+10c\left(2\right)\)

\(n_{H^+}=4n_{NO}+2n_{NO_2}+12n_{N_2}\)

\(\Rightarrow4a+2b+12c=1.2\left(3\right)\)

\(\left(1\right),\left(2\right),\left(3\right):a=b=0.1,c=0.05\)

\(\%V_{NO_2}=\dfrac{0.1}{0.25}\cdot100\%=40\%\%0-\)

\(m_X=56x+64y+108x=29.2\left(g\right)\)

\(\Rightarrow164x+64y=29.2\left(1\right)\)

\(n_{H_2SO_4}=\dfrac{80\cdot98\%}{98}=0.8\left(mol\right)\)

Bảo toàn e : 

\(3n_{Fe}+2n_{Cu}+n_{Ag}=2n_{SO_2}\)

\(\Rightarrow n_{SO_2}=2x+y\left(mol\right)\)

Bảo toàn S : 

\(n_{H_2SO_4}=3n_{Fe_2\left(SO_4\right)_3}+n_{CuSO_4}+n_{Ag_2SO_4}+n_{SO_2}\)

\(\Rightarrow x+y+x+2x+y=0.8\)

\(\Rightarrow4x+2y=0.8\left(2\right)\)

\(\left(1\right),\left(2\right):x=0.1,y=0.2\)

Chất rắn không tan là Cu chưa pư.

⇒ m_{Cu (dư)} = 6 (g)

Ta có: 56n_Fe + 64n_{Cu (pư)} = 30 - 6 (1)

\(n_{HNO_3}=0,52.2=1,04\left(g\right)\Rightarrow n_{NO}=\dfrac{1,04}{4}=0,26\left(mol\right)\)

BT e, có: 2n_Fe + 2n_{Cu (pư)} = 3n_NO = 0,78(2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,12\left(mol\right)\\n_{Cu\left(pư\right)}=0,27\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Cu}=\dfrac{0,27.64+6}{30}.100\%=77,6\%\)

Ta có: 24n_Mg + 27n_Al = 15,12 (1)

\(n_{HNO_3}=0,9.2=1,8\left(mol\right)\)

Mà: n_HNO3 = 10n_N2O ⇒ n_N2O = 0,18 (mol)

BT e, có: 2n_Mg + 3n_Al = 8n_N2O = 1,44 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,36\left(mol\right)\\n_{Al}=0,24\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Mg}=\dfrac{0,36.24}{15,12}.100\%\approx57,14\%\)

\(a.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ b.n_{H_2SO_4}=0,22.1,25=0,275mol\\ n_{Fe_2O_3}=a;n_{CuO}=b\\ \Rightarrow\left\{{}\begin{matrix}3a+b=0,275\\160a+80b=16\end{matrix}\right.\\ \Rightarrow a=0,075;b=0,05\\ \%m_{Fe_2O_3}=\dfrac{0,075.160}{16}\cdot100=75\%\\ \%m_{CuO}=100-75=25\%\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ n_{Al}=a;n_{Mg}=b\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow\left\{{}\begin{matrix}1,5a+b=0,4\\27a+24b=7,8\end{matrix}\right.\\ \Rightarrow a=0,2;b=0,1\\ \%m_{Al}=\dfrac{0,2.27}{7,8}\cdot100=69,23\%\\ \%m_{Mg}=100-69,23=30,77\%\)

cảm ơn b nha này đánh giá 5 sao dc hong z