mẫu giấy quỳ tím hóa đỏ sau đó mất màu do có HClO có tính tẩy màu

$16HCl +2KMnO_{4} --> 2KCl +2MnCl_{2} +5Cl_{2}+8H_{2}O$

$Cl_{2} +H_{2}O --> HCl +HClO$

16HCl+2KMnO4−−>2KCl+2MnCl2+5Cl2+8H2O

Cl2+H2O−−>HCl+HClO

mẫu giấy quỳ tím hóa đỏ sau đó mất màu do có HClO có tính tẩy màu

Gọi số mol NaF là x suy ra NaCl là 2x.

\(\Rightarrow m_{hh}=m_{NaCl}+m_{NaF}=2x.58,5+x.42=15,9\)

\(\Rightarrow x=0,1\)

Dẫn hỗn hợp vào dung dịch X

\(NaCl+AgNO_3\rightarrow AgCl+NaNO_3\)

\(\Rightarrow n_{AgCl}=n_{NaCl}=0,2\left(mol\right)\)

\(\Rightarrow m_{AgCl}=0,2.\left(108+35,5\right)=28,7\left(g\right)\)

\(\rightarrow m_Y=28,7-15,9=12,8\left(g\right)\)

PTPU

NaCl+ AgNO₃\(\rightarrow\) NaNO₃+ AgCl\(\downarrow\)

.2a...........2a...........................2a...

có: 58,5. 2a+ 42a= 15,9

\(\Rightarrow\) a= 0,1( mol)

\(\Rightarrow\) n_NaCl= 0,2( mol)

có: m_{dd giảm}= m_{AgCl\(\downarrow\)}

= 0,2. 143,5= 28,7( g)

Ancol no này có dạng

Khối lượng X tăng là do lấy O từ CuO.

\(\Rightarrow m_O=6,2-4,6=1,6\left(g\right)\Rightarrow n_O=\frac{1,6}{16}=0,1\left(mol\right)=n_{Andehit}< n_{Ancol}\)

\(\)\(\Rightarrow M_{Ancol}< \frac{4,6}{0,1}=46\)

Vậy ancol là suy ra andehit là HCHO (tráng 4Ag)

\(\Rightarrow n_{Ag}=4n_{HCHO}=4.0,1=0,4\left(mol\right)\)

\(\Rightarrow m_{Ag}=m=0,4.108=43,2\left(g\right)\)

RCH2OH + [O] → RCHO + H2O

Bảo toàn khối lượng: mancol + mO = mhỗn hợp X => nO phản ứng = (6,2 – 4,6) / 16 = 0,1 mol

=> nRCH2OH phản ứng = 0,1 mol => nRCH2OH ban đầu > 0,1 mol

=> Mancol < 4,6 / 0,1 = 46 => ancol là CH3OH => anđehit tạo ra là HCHO

nAg = 4.nHCHO = 0,4 mol => mAg = 43,2 gam

Phản ứng xảy ra:

\(2H_2+O_2\rightarrow2H_2O\)

\(m_{H2SO4}=54.96\%=51,84\left(g\right)\)

\(\Rightarrow m_{ddH2SO4\left(72\%\right)}=\frac{51,84}{72\%}=72\left(g\right)\)

\(\Rightarrow m_{H2O\left(them.vao\right)}=72-54=18\left(g\right)\)

\(n_{H2O}=\frac{18}{18}=1\left(mol\right)\)

\(\Rightarrow n_{H2}=n_{H2O}=1\left(mol\right);n_{O2}=\frac{1}{2}n_{H2O}=0,5\left(mol\right)\)

\(\Rightarrow n_X=1+0,5=1,5\left(mol\right)\)

\(\Rightarrow V=1,5.22,4=33,6\left(l\right)\)

\(m_{dd\left(tang\right)}=m_{H2\left(mat.di\right)}=5.2-4,9=0,3\left(mol\right)\)

\(\Rightarrow n_{H2}=0,15\left(mol\right)\Rightarrow n_{HCl}=2.0,15=0,3\left(mol\right)\)

BTKL:

\(m_{hh}=m_{HCl}+m_{muoi}+m_{H2}\)

\(\Rightarrow m_{muoi}=5,2-0,5.36,5-0,3=15,85\left(g\right)\)

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

0,055--------------------0,0825 mol

nH2=1,848\22,4=0,0825 mol

=>mAl=0,055.27=1,485 g

=>%mAl=1,485\3,75.100=39,6 %

=>%mAl2O3=100-39,6=60,4 %

a) \(2Al+6HCl-->2AlCl3+3H2\)

\(Al2O3+6HCl-->2AlCl3+3H2O\)

\(n_{H2}=\frac{1,848}{22,4}=0,0825\left(mol\right)\)

\(n_{Al}=\frac{2}{3}n_{H2}=0,055\left(mol\right)\)

\(\%m_{Al}=\frac{0,055.27}{3,72}.100\%=40\%\)

\(\%m_{Al2O3}=100-40=60\%\)

b) \(n_{HCl\left(1\right)}=2n_{H2}=0,165\left(mol\right)\)

\(m_{Al2O3}=3,72-0,055.27=2,235\left(mol\right)\)

\(n_{Al2O3}=\frac{2,235}{102}=0,022\left(mol\right)\)

\(n_{HCl}=6n_{Al2O3}=0,132\left(mol\right)\)

\(\sum n_{HCl}=0,132+0,165=0,297\left(mol\right)\)

\(V_{HCl}=\frac{416}{1,04}=400ml=0,4\left(l\right)\)

\(C_{M\left(HCl\right)}=\frac{0,297}{0,4}=0,7425\left(M\right)\)

\(m_{dd}=m_{hh}+m_{ddHCl}-m_{H2}=3,72+416-0,165=419,555\left(g\right)\)

\(m_{AlCl3\left(1\right)}=0,055.133,5=7,3425\left(g\right)\)

\(m_{AlCl3\left(2\right)}=0,044.133,5=5,874\left(g\right)\)

\(\sum m_{AlCl3}=5,874+7,3425=13,2165\left(g\right)\)

\(C\%_{AlCl3}=\frac{13,2165}{419,555}.100\%=3,15\%\)

Phản ứng xảy ra:

\(n_{KClO3}=\frac{24,5}{39+35,5+16.3}=0,2\left(mol\right)\)

\(\Rightarrow n_{KClO2\left(pư\right)}=0,2.70\%=0,14\left(mol\right)\)

\(\Rightarrow n_{Cl2}=0,14.3=0,42\left(mol\right)\)

\(2NaOH+Cl_2\rightarrow NaCl+NaClO\)

\(\Rightarrow n_{NaCl}=n_{NaClO}=n_{Cl2}=0,42\left(mol\right)\)

\(\Rightarrow m=m_{NaCl}+m_{NaClO}=0,42.58,5+0,42.74,5=55,86\left(g\right)\)

-----------0,2----------0,6 mol

Cl2+2NaOH-->NaCl+NaClO+H2O

0,6-----------------0,6-----0,6 mol

nKClO3=24,5\122,5=0,2 mol

=>VCl2=0,6.22,4=13,44 l

=>mChất tan= 0,6.58,5+0,6.74,5=79,8 g

ko cho H2 là bao nhiêu ạ

ko bạn ơi

\(MnO2+4HCl--.Cl2+2H2O+MnCl2\)

\(n_{Cl2}=\frac{7,84}{22,4}=0,35\left(mol\right)\)

\(nMnO2=n_{Cl2}=0,35\left(mol\right)\)

\(m=m_{MnO2}=0,35.87=30,45\left(g\right)\)

MnO2+4HCl−−>MnCl2+Cl2+2H2O

0,35---------------------------0,35 mol

nCl2=7,84\22,4=0,35(mol)

m=mMnO2=0,35.87=30,45(g)

a)\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(Fe+2HCl--.FeCl2+H2\)

x-----------------------------------x(mol)

\(Mg+2HCl-->MgCl2+H2\)

y-------------------------------------y(mol)

theo bài ta có hpt

\(\left\{{}\begin{matrix}56x+24y=9,2\\x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)

\(\%m_{Fe}=\frac{0,1.56}{9,2}.100\%=60,87\%\)

\(\%m_{Mg}=100-60,87=39,13\%\)

b) \(n_{HCl}=2n_{H2}=0,5\left(mol\right)\)

\(V_{HCl}=V_{dd}=\frac{0,5}{2}=0,25\left(l\right)\)

\(C_{M\left(FeCl2\right)}=\frac{0,1}{0,25}=0,4\left(M\right)\)

\(C_{M\left(Mgcl2\right)}=\frac{0,15}{0,25}=0,6\left(M\right)\)