\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

               1<------------------------------------0,5

=> \(m_{KMnO_4\left(pthh\right)}=1.158=158\left(g\right)\)

=> \(m_{KMnO_4\left(tt\right)}=\dfrac{158.100}{80}=197,5\left(g\right)\)

PT: \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

Ta có: \(n_{Cl_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{KMnO_4\left(LT\right)}=\dfrac{2}{5}n_{Cl_2}=0,2\left(mol\right)\\n_{HCl\left(LT\right)}=\dfrac{16}{5}n_{Cl_2}=1,6\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{KMnO_4\left(LT\right)}=0,2.158=31,6\left(g\right)\\V_{ddHCl\left(LT\right)}=\dfrac{1,6}{2}=0,8\left(l\right)\end{matrix}\right.\)

Mà: H% = 75%

\(\Rightarrow\left\{{}\begin{matrix}m_{KMnO_4\left(TT\right)}=\dfrac{31,6}{75\%}\approx42,13\left(g\right)\\V_{ddHCl\left(TT\right)}=\dfrac{0,8}{75\%}\approx1,067\left(l\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

Chọn đáp án C

\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{47,4}{158}=0,3mol\)

\(n_{KMnO_4}=\dfrac{0,3}{80\%}=0,375mol\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

     2                16             2            2                5       8         ( mol )

 0,375      >      2,5                                                     ( mol )

 0,375                                                             0,9375   ( mol )

\(V_{Cl_2}=n_{Cl_2}.22,4=0,9375.22,4=21l\)

\(n_{KMnO_4\left(bd\right)}=\dfrac{47,4}{158}=0,3\left(mol\right)\) => \(n_{KMnO_4\left(pư\right)}=\dfrac{0,3.80}{100}=0,24\left(mol\right)\)

PTHH: 2KMnO₄ + 16HCl --> 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

            0,24------------------------------------->0,6

=> \(V=0,6.22,4=13,44\left(l\right)\)

Câu 2: 

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

               0,6<------------------------------------0,3

\(\Rightarrow m_{KMnO_4}=0,6.158=94,8\left(g\right)\)

Câu 3: 

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

             0,2<-----------------------0,2

=> m_Zn = 0,2.65 = 13 (g)

Câu 4:

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

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

           0,4------------------------->0,4

             \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

              0,4<---0,4

\(\Rightarrow m_{CuO}=0,4.80=32\left(g\right)\)

a, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,2}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

b, \(n_{H_2O}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,3.18=5,4\left(g\right)\)

c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

_______0,3_______________________0,15 (mol)

\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)

Bạn tham khảo nhé!

Ta có nCl2 = 8,96/22,4 = 0,4 mol

PTHH :

2KMnO4 + 16HCl - > 2KCl + 2MnCl2 + 5Cl2 + 8H2O

0,16mol.........1,28mol...............................0,4mol

=> Khối lượng của KMnO4 là : mKMnO4 = 0,16.158=25,28(g)

Khối lượng dd HCl là : mddHCl = \(\frac{1,28.36,5.100}{19,2}\approx243,33\left(g\right)\)

Vì hiệu suất là 80% nên

=> \(\left\{{}\begin{matrix}mKMnO4=\frac{25,28.80}{100}=20,224\left(g\right)\\mddHCl=\frac{243,33.80}{100}=194,664\left(g\right)\end{matrix}\right.\)

a) $n_{O_2} = 0,15(mol)$

\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

0,3                         0,15           0,15       0,15                    (mol)

$H = \dfrac{0,15.158}{63,2}.100\% = 37,5\%$

b)

$m_B = 63,2 - 0,15.32 = 58,4(gam)$
$\%m_{K_2MnO_4} = \dfrac{0,15.197}{58,4}.100\% = 50,59\%$

$\%m_{MnO_2} = \dfrac{0,15.87}{58,4}.100\% = 22,35\%$
$\%m_{KMnO_4\ dư} = 100\% -50,59\% -22,35\% = 27,06\%$

a. PTHH: \(KMnO_4\rightarrow^{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

b. \(H=100\%\)

\(n_{KMnO_4}=\frac{3,6}{158}=0,023mol\)

Theo phương trình \(n_{O_2}=0,5n_{KMnO_4}=0,046mol\)

\(\rightarrow V_{O_2}=0,0115.22,4.100\%=0,2576l\)

c. H = 80%

\(\rightarrow V_{O_2}=0,0115.22,4.80\%=0,20608l\)