\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ n_{Fe}=a;n_{Al}=b\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}56a+27b=11\\a+1,5b=0,4\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\)

\(\%m_{Fe}=\dfrac{0,1.56}{11}\cdot100=50,91\%\\ \%m_{Al}=100-50,91=49,09\%\)

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

0,1            0,2          0,1          0,1

\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)

0,2          0,6           0,2            0,3

\(m_{HCl}=\dfrac{\left(0,2+0,6\right).36,5}{9,125}\cdot100=320g\)       

\(c)m_{dd}=320+11-0,1.2-0,3.2=308,2g\)

\(C_{\%FeCl_2}=\dfrac{0,1.127}{308,2}\cdot100=4,12\%\\ C_{\%AlCl_3}=\dfrac{0,2.133,5}{308,2}\cdot100=8,66\%\)

\(a)n_{HCl}=0,1.3=0,3mol\\ n_{CuO}=a;n_{ZnO}=b\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ \Rightarrow\left[{}\begin{matrix}80a+81b=12,1\\2a+2b=0,3\end{matrix}\right.\\ \Rightarrow a=0,05;b=0,1\\ \%m_{CuO}=\dfrac{80.0,05}{12,1}\cdot100=33,06\%\\ \%m_{ZnO}=100-33,06=66,94\%\\ b)ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O \\ n_{H_2SO_4}=0,05+0,1=0,15mol\\ m_{ddH_2SO_4}=\dfrac{0,15.98}{20}\cdot100=73,5g\)

bạn đổi ngoặc [ thành ngoặc { giúp mik nhé!

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=n_{H_2}=0,3mol\\ m_{rắn}=0,3.56+2,4=19,2g\)

a, Ta có: \(n_{KOH}=0,15.2=0,3\left(mol\right)\)

PT: \(KOH+HCl\rightarrow KCl+H_2O\)

Theo PT: \(n_{HCl\left(pư\right)}=n_{KOH}=0,3\left(mol\right)\)

Mà: HCl dư 15%

\(\Rightarrow n_{HCl}=0,3+0,3.15\%=0,345\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,345}{1,5}=0,23\left(l\right)\)

b, Theo PT: \(n_{KCl}=n_{KOH}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{KCl}}=\dfrac{0,3}{0,15+0,23}\approx0,789\left(M\right)\)

\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,3.15\%}{0,15+0,23}\approx0,118\left(M\right)\)

Ta có: \(m_{Ba\left(OH\right)_2}=100.17,1\%=17,1\left(g\right)\Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{17,1}{171}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=150.9,8\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

PT: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_{4\downarrow}+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,15-0,1=0,05\left(mol\right)\)

Ta có: m _{dd sau pư} = 100 + 150 - 0,1.233 = 226,7 (g)

\(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,05.98}{226,7}.100\%\approx2,16\%\)

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

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

______0,05_____0,1______0,05 (mol)

\(\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{3,65}{7,3\%}=50\left(g\right)\)

Ta có: m _{dd sau pư} = 4 + 50 = 54 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{0,05.135}{54}.100\%=12,5\%\)

Đề cho 4 (g) chất gì bạn nhỉ?

a, Ta có: \(m_{H_2SO_4}=500.5,88\%=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

______0,2________0,3_______0,1______0,3 (mol)

\(m_{Al}=0,2.27=5,4\left(g\right)\)

\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b, Ta có: m _{dd sau pư} = 5,4 + 500 - 0,3.2 = 504,8 (g)

\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{504,8}.100\%\approx6,77\%\)

\(a)n_{H_2SO_4}=\dfrac{500.5,88}{100.98}=0,3mol\\2 Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2          0,3               0,1                0,3

\(m_{Al}=0,2.27=5,4g\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72l\\ b)C_{\%Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{500+5,4-0,3.2}\cdot100=6,77\%\)

\(a)n_{H_2SO_4}=0,15.1=0,15mol\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,3                 0,15            0,15              0,15

\(C_{M\left(NaOH\right)}=\dfrac{0,3}{0,1}=3M\\ b)C_{M\left(Na_2SO_4\right)}=\dfrac{0,15}{0,1+0,15}=0,6M\)