\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

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

\(0.1......0.2....................0.1\)

\(m_{Cu}=m_{hh}-m_{Fe}=12-0.1\cdot56=6.4\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,03.24=0,72\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,72}{1,74}.100\%\approx41,38\%\\\%m_{AlCl_3}\approx58,62\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,06.36,5=2,19\left(g\right)\)

\(\Rightarrow C\%_{ddHCl}=\dfrac{2,19}{500}.100\%=0,438\%\)

Bạn tham khảo nhé!

a, nH2 = 0,03 ( mol )

=> nMg = nH2 = 0,03 ( mol )

=> mMg = 0,72 g

=> %Mg \(\approx\) 41,38 % .

=> % Al \(\approx\) 58,62 % .

b, Có : nH2 = 0,03 mol

=> nHCl = nHCl_{từ Al2O3} + nHCl_{từ Mg} = 0,06 + 0,06 = 0,12 ( mol )

=> mHCl = 4,38 ( g )

Lại có : mdd = mhh + mddHCl = 501,74 ( g )

=> \(C\%=\dfrac{m_{HCl}}{m_{dd}}.100\%\approx0,87\%\)

( chắc đoạn trên là Al2O3 :vvvv )

a/ Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

n_H2 = \(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PTHH: n_H2 = n_Fe = 0,15 (mol) \(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

\(\Rightarrow m_{Cu}=11-8,4=2,6\left(g\right)\)

\(\Rightarrow\%m_{Fe}=\dfrac{8,4}{11}.100\%\approx76,4\%\)

\(\Rightarrow\%m_{Cu}=100-76,4\approx23,6\%\)

b/ Theo PTHH ta có: n_HCl = 2n_Fe = 2.0,15 = 0,3 (mol)

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(M\right)\)

c/ m_HCl = 36,5 . 0,3 = 10,95(g)

\(\Rightarrow C\%_{HCl}=\dfrac{m_{HCl}}{m_{ddHCl}}.100\%=\dfrac{10,95}{200}.100\%=5,475\%\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{11}.100\%\approx76,36\%\\\%m_{Cu}\approx23,64\%\end{matrix}\right.\)

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

\(\Rightarrow V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)

c, \(C\%_{HCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)

PTHH\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)

tl............1................2.............2.............1.............1..(mol

br     0,1.................0,2......................................0,1(mol)

NaCl không phản ứng đc vsHCl

b)\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(C_{MHCl}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)(đổi 400ml=0,4(l))

c)\(Tacom_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{10,6}{20}.100=53\%\)

\(\Rightarrow\%mNaCl=100\%-53\%=47\%\)