dd chi đó bn

\(1.Na_2O+H_2O\rightarrow2NaOH\\ 2.m_{rắn}=m_{CuO}=6,9g\\ m_{Na_2O}=10-6,9=3,1g\\ n_{Na_2O}=\dfrac{3,1}{62}=0,05mol\\ n_{NaOH}=0,05.2=0,1mol\\ 200ml=0,2l\\ C_{M_X}=C_{M_{NaOH}}=\dfrac{0,1}{0,2}=0,5M\)

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

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

b, Ta có: 80n_CuO + 81n_ZnO = 12,1 (1)

Theo PT: \(n_{HCl}=2n_{CuO}+2n_{ZnO}=0,2.1,5=0,3\left(mol\right)\left(2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CuO}=0,05\left(mol\right)\\n_{ZnO}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{12,1}.100\%\approx33,06\%\\\%m_{ZnO}\approx66,94\%\end{matrix}\right.\)

a, PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{FeO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 72y = 11,2 (1)

Ta có: \(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}+n_{FeO}=x+y=0,15\left(2\right)\)

Từ (1) và (2) ⇒ x = 0,05 (mol), y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{11,2}.100\%\approx35,71\%\\\%m_{FeO}\approx64,28\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{CuSO_4}=n_{Cu}=0,05\left(mol\right)\\n_{FeSO_4}=n_{FeO}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{CuSO_4}}=\dfrac{0,05}{0,15}=\dfrac{1}{3}\left(M\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\end{matrix}\right.\)

\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)

\(m=80a+160b=20\left(g\right)\left(1\right)\)

\(n_{HCl}=0.2\cdot3.5=0.7\left(mol\right)\)

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

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(n_{HCl}=2a+6b=0.7\left(mol\right)\left(2\right)\)

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

\(m_{CuO}=0.05\cdot80=4\left(g\right)\)

\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)

\(n_{FeO}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)

\(m_{hh}=72a+80b=19.2\left(g\right)\left(1\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(n_{H_2SO_4}=a+b=0.25\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.15\)

\(m_{FeO}=0.1\cdot72=7.2\left(g\right)\)

\(m_{CuO}=12\left(g\right)\)

\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)

\(C_{M_{CuSO_4}}=\dfrac{0.15}{0.25}=0.6\left(M\right)\)

CuO + 2HCl -> CuCl₂ + H₂O (1)

Fe₂O₃ + 6HCl -> 2FeCl₃ + 3H₂O (2)

n_HCl=0,2.3,5=0,7(mol)

Đặt n_CuO=a

n_Fe2O3=b

Ta có hệ:

80a+160b=20

2a+6b=0,7

=>a=0,05;b=0,1

m_CuO=80.0,05=4(g)

m_Fe2O3=20-4=16(g)

Theo PTHH 1 và 2 ta có:

n_CuCl2=n_CuO=0,05(mol)

n_FeCl3=2n_Fe2O3=0,2(mol)

m_CuCl2=135.0,05=6,75(g)

m_FeCl3=162,5.0,2=32,5(g)

m_dd =20+200.1,1=240(g)

C% dd CuCl₂=6,72\240 .100%=2,8125%

C% dd FeCl₃= 32,5\240 .100%=13,54%

Câu 3:

Gọi x, y lần lượt là số mol của MgO và Al₂O₃

Ta có: \(n_{H_2SO_4}=0,2.250:1000=0,05\left(mol\right)\)

a. PTHH: 

MgO + H₂SO₄ ---> MgSO₄ + H₂O (1)

Al₂O₃ + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂O (2)

b. Theo PT₍₁₎: \(n_{H_2SO_4}=n_{MgO}=x\left(mol\right)\)

Theo PT₍₂₎: \(n_{H_2SO_4}=3.n_{Al_2O_3}=3y\left(mol\right)\)

=> x + 3y = 0,05 (1)

Theo đề, ta có: 40x + 102y = 1,82 (2)

Từ (1) và (2), ta có HPT:

\(\left\{{}\begin{matrix}x+3y=0,05\\40x+102y=1,82\end{matrix}\right.\)

=> x = 0,02, y = 0,01

=> \(m_{MgO}=0,02.40=0,8\left(mol\right)\)

=> \(\%_{m_{MgO}}=\dfrac{0,8}{1,82}.100\%=43,96\%\)

\(\%_{m_{Al_2O_3}}=100\%-43,96\%=56,04\%\)

Câu 4:

Ta có: \(m_{H_2SO_4}=\dfrac{19,6\%.100\%}{100}=19,6\left(g\right)\)

=> \(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Ta lại có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

a. PTHH: CuO + H₂SO₄ ---> CuSO₄ + H₂O

Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\)

Vậy H₂SO₄ dư.

Theo PT: \(n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

Ta có: \(m_{dd_{CuSO_4}}=8+100=108\left(g\right)\)

=> \(C_{\%_{CuSO_4}}=\dfrac{16}{108}.100\%=14,81\%\)

Câu 5: Thiếu đề