Bài 1: a) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

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

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

=> \(m_{ddHCl}=\dfrac{0,6.36,5}{14,6\%}=150\left(g\right)\)

b) \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

=> \(m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

Theo PT: \(n_{H_2}=n_{H_2SO_4}=\dfrac{4,9\%.100}{98}=0,05\left(mol\right)\)

=> \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b)Theo PT: \(n_{Mg}=n_{H_2SO_4}=0,05\left(mol\right)\)

=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)

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

a)\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT:  \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

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

b)Theo PT:  \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

=> \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

c)Theo PT:  \(n_{HCl}=3n_{Al}=0,6\left(mol\right)\)

=> \(CM_{HCl}=\dfrac{0,6}{0,2}=3M\)

giup mk voi

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

0,2     0,2            0,2           0,2

a)\(V_{H_2}=0,2\cdot22,4=4,48l\)

b)\(m_{ZnSO_4}=0,2\cdot161=32,2g\)

  \(m_{ddZnSO_4}=30+200-0,2\cdot2=229,6g\)

  \(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{32,2}{229,6}\cdot100\%=14,02\%\)

c)\(n_{CuO}=\dfrac{24}{80}=0,3mol\)

   \(CuO+H_2\rightarrow Cu+H_2O\)

   0,3        0,2     0,2

   \(m_{rắn}=m_{Cu}=0,2\cdot64=12,8g\)

C%=mctmdd⋅100%=19,6200⋅100%=9,8%C%=mctmdd⋅100%=19,6200⋅100%=9,8%

c)

bn mik hỏi tí aj
đoạn mdd ZnSO4 ấy, 30 ở đâu ra v ạ

Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

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

______0,2_____0,4_____0,2 (mol)

a, \(m_{CuCl_2}=0,2.135=27\left(g\right)\)

b, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{14,6}{300}.100\%\approx4,867\%\)

c, Ta có: m _{dd sau pư} = 16 + 300 = 316 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{27}{316}.100\%\approx8,54\%\)

a)

Khối lượng của dung dịch:

\(m_{dd}=m_{ct}+m_{dm}=20+180=200\left(g\right)\)

Nồng độ phần trăm của dung dịch:

\(C\%=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{20}{200}.100\%=10\%\)

b) đề sai nha bạn

a) \(n_{FeCl_3}=\dfrac{16,25}{162,5}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

            0,05<----0,3<-----0,1

=> \(m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b) 

\(m_{HCl\left(bd\right)}=91,25.16\%=14,6\left(g\right)\)

m_{dd sau pư} = 8 + 91,25 = 99,25 (g)

\(\left\{{}\begin{matrix}C\%\left(FeCl_3\right)=\dfrac{16,25}{99,25}.100\%=16,373\%\\C\%\left(HCldư\right)=\dfrac{14,6-0,3.36,5}{99,25}.100\%=3,678\%\end{matrix}\right.\)

Đáp án là D

\(n_{HCl}=1\cdot0,2=0,2\left(mol\right)\\ PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\\ a,n_{MgO}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m=m_{MgO}=0,1\cdot40=4\left(g\right)\\ b,n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\ c,m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{250}\cdot100\%=2,92\%\)

\(n_{H_2O}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{H_2O}=0,1\cdot18=1,8\left(g\right)\\ \Rightarrow m_{dd_{MgCl_2}}=4+250-1,8=252,2\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{252,2}\cdot100\%\approx3,77\%\)