\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

 0,4           0,4         0,4        0,4

a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\) 

 \(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)

b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)

   \(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)

   \(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)

c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)

     \(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)

         0,4          0,3         0,3              0,3

     \(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

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

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)

Giúp với mai mình thi rồi!

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)

b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

c, \(C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Màu đỏ nha :))

nAl= 0,5(mol)

a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2

nHCl= 6/2 . 0,5= 1,5(mol)

=>mHCl= 1,5.36,5=54,75(mol)

=> mddHCl= (54,75.100)/18,25=300(g)

b) nH2= 3/2. 0,5=0,75(mol)

=>V(H2,đktc)=0,75.22,4=16,8(l)

c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)

mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)

=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)

a, Gọi CTHH cần tìm là R₂O_n.

PT: \(R_2O_n+2nHCl\rightarrow2RCl_n+nH_2O\)

Ta có: \(n_{R_2O_n}=\dfrac{5,6}{2M_R+16n}\left(mol\right)\)

\(n_{RCl_n}=\dfrac{11,1}{M_R+35,5n}\left(mol\right)\)

Theo PT: \(n_{RCl_n}=2n_{R_2O_n}\Rightarrow\dfrac{11,1}{M_R+35,5n}=\dfrac{2.5,6}{2M_R+16n}\)

\(\Rightarrow M_R=20n\)

Với n = 2 thì M_R = 40 (g/mol) là thỏa mãn.

→ CaO.

b, PT: \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

_______0,1_____0,2 (mol)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{17,8\%}\approx41,011\left(g\right)\)

Ta có: m _{dd sau pư} = 5,6 + 41,011 = 46,611 (g)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{11,1}{46,611}.100\%\approx23,81\%\)

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

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

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,2        0,4           0,2            0,2

\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)