\(1.Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO\uparrow+2H_2O\)

\(2.4Al\left(NO_3\right)_3\underrightarrow{t^0}2Al_2O_3+12NO\uparrow+3O_2\)

\(3.2Al_2O_3\underrightarrow{đpnc}4Al+3O_2\)

\(4.2Al+Ba\left(OH\right)_2+2H_2O\rightarrow Ba\left(AlO_2\right)_2+3H_2\)

3 thêm xt criolic nx

\(a,n_{HCl}=\dfrac{120.19,6}{100.36,5}=\dfrac{1176}{1825}mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{1176}{1825}:2=\dfrac{588}{1825}mol\\ m_{Fe}=\dfrac{588}{1825}\cdot56\approx18,04g\\ b,V_{H_2}=\dfrac{588}{1825}\cdot22,4\approx7,22l\)

Thầy thấy tỉ lệ 19,6% sẽ đẹp số mol khi đó là dung dịch H₂SO₄, em kiểm tra lại đề nhé!

Áp dụng định luật bảo toàn Clo, ta được:

\(\%Cl\left(pư\right)=\%Cl\left(sp\right)\)

=>\(32,85\%=24,2\%+\left(2a\right)\%\)

=>\(32,85=24,2+a\cdot2\)

=>\(a\cdot2=32,85-24,2=8,65\)

=>\(a=\dfrac{8.65}{2}=4.325\)

Áp dụng định luật bảo toàn Clo, ta được:

$\%Cl\left(pư\right)=\%Cl\left(sp\right)$

=>$32,85\%=24,2\%+\left(2a\right)\%$

=>$32,85=24,2+a\cdot 2$

=>$a\cdot 2=32,85-24,2=8,65$

=>$a=\frac{8.65}{2}=4.325$

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

Theo ĐLBT KL: m_Zn + m_H2SO4 = m_ZnSO4 + m_H2

⇒ m_H2SO4 = 9 + 0,1.2 - 6,5 = 2,7 (g)

\(a.500ml=0,5l\\ C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,5}=0,2M\\ b.m_{ddNa_2CO_3}=500.1=500g\\ m_{Na_2CO_3}=0,1.106=10,6g\\ C_{\%Na_2CO_3}=\dfrac{10,6}{500}\cdot100\%=2,12\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{20\%}==73\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,2}{2,5}=0,08\left(l\right)\)

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

a, \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Mg}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a, \(n_{H_2}=n_{Zn}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)

b, \(n_{HCl}=2n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)

c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)

\(Fe_2O_3+3CO\rightarrow\left(t^o\right)2Fe+3CO_2\\ n_{CO}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ n_{Fe_2O_3}=\dfrac{0,3}{3}=0,1\left(mol\right);n_{Fe}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ m_{Fe}=0,2.56=11,2\left(g\right)\\ m_{Fe_2O_3}=0,1.160=16\left(g\right)\)