a) Sau phản ứng : $m_{chất\ rắn} = 18,88(gam)$

b) Bảo toàn khối lượng : 

$m_{O_2} = 20 - 18,8 = 1,12(gam)$

$n_{O_2} = 1,12 : 32 = 0,035(mol)$
$V_{O_2} = 0,035.22,4 = 0,784(lít)$

Gọi n KMnO4 = a 

       n KClO3 = b ( mol ) 

--> 158a + 122,5 b = 43,3 

PTHH :

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)

0,9b                       1,35b

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

0,9a                                             0,45a 

\(\%Mn=\dfrac{55a}{43,3-32\left(0,45a+1,35b\right)}=24,103\%\)

\(\rightarrow a=0,15\)

\(b=0,16\)

\(m_{KMnO_4}=0,15.158=23,7\left(g\right)\)

\(m_{KClO_3}=0,16.122,5=19,6\left(g\right)\)

\(a,n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:2K+2H_2O\rightarrow2KOH+H_2\uparrow\\ Theo.pt:n_K=2n_{H_2}=2.0,1=0,2\left(mol\right)\\ m_K=0,2.39=7,8\left(g\right)\\ m_{K_2O}=17,2-7,8=9,4\left(g\right)\\ b,n_{CuO\left(bđ\right)}=\dfrac{12}{80}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:0,15>0,1\Rightarrow Cu.dư\)

Gọi n_{CuO (pư)} = a (mol)

=> n_Cu = a (mol)

m_{chất rắn sau pư} = 80(0,15 - a) + 64a = 10,8

=> a = 0,075 (mol)

=> n_{H2 (pư)} = 0,075 (mol)

\(H=\dfrac{0,075}{0,1}=75\%\)

Kudo làm sai , anh làm lại nha!

\(a,n_{BaCO_3}=\dfrac{49,25}{197}=0,25\left(mol\right)\\ PTHH:BaCO_3\rightarrow\left(t^o\right)BaO+CO_2\\ n_{CO_2\left(LT\right)}=n_{BaO\left(LT\right)}=n_{BaCO_3}=0,25\left(mol\right)\\ \Rightarrow n_{BaO\left(TT\right)}=n_{CO_2\left(TT\right)}=0,6.0,25=0,15\left(mol\right)\\ V_{CO_2\left(TT,đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,BaCO_3\rightarrow\left(t^o\right)BaO+CO_2\\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{BaCO_3\left(LT\right)}=n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow n_{BaCO_3\left(TT\right)}=\dfrac{0,15}{80\%}=0,1875\left(mol\right)\\ m_{BaCO_3\left(TT\right)}=a=0,1875.197=\dfrac{591}{16}\left(g\right)\\ c,n_{BaCO_3}=\dfrac{49,25}{197}=0,25\left(mol\right)\)

\(Đặt:n_{BaCO_3\left(tg\right)}=j\left(mol\right)\left(a>0\right) \\Mà:m_{rắn}=42,65\\ \Leftrightarrow137j+\left(49,25-197j\right)=42,65\\ \Leftrightarrow j=11\%\\ \Rightarrow H=\dfrac{j}{0,25}.100\%=\dfrac{0,11}{0,25}.100=44\%\)

C lỗi đề nha bạn

1) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

m_A = m_KMnO4(bđ) - m_O2 = 79 - 0,15.32 = 74,2 (g)

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

               0,3<-----------0,15<----0,15<---0,15

=> \(H=\dfrac{0,3.158}{79}.100\%=60\%\)

2) 

\(\left\{{}\begin{matrix}\%m_{K_2MnO_4}=\dfrac{0,15.197}{74,2}.100\%=39,825\%\\\%m_{MnO_2}=\dfrac{0,15.87}{74,2}.100\%=17,588\%\\\%m_{KMnO_4\left(không.pư\right)}=\dfrac{79-0,3.158}{74,2}.100\%=42,587\%\end{matrix}\right.\)

3) \(n_{KMnO_4\left(không.pư\right)}=\dfrac{79}{158}-0,3=0,2\left(mol\right)\)

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

                 0,2----------------------------------->0,5

             K₂MnO₄ + 8HCl --> 2KCl + MnCl₂ + 2Cl₂ + 4H₂O

                 0,15-------------------------------->0,3

              MnO₂ + 4Hcl --> MnCl₂ + Cl₂ + 2H₂O

                0,15------------------->0,15

=> \(V_{Cl_2}=22,4\left(0,5+0,3+0,15\right)=21,28\left(l\right)\)

\(n_{KMnO_4}=\dfrac{79}{158}=0,5mol\)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

0,5                                                  0,15

a)\(m_{KMnO_4}=0,15\cdot197=29,55g\)

\(m_{MnO_2}=0,15\cdot87=13,05g\)

\(m_{CRắn}=m_{KMnO_4}+m_{MnO_2}=29,55+13,05=42,6g\)

\(n_{KMnO_4pư}=0,15\cdot2=0,3mol\)

\(H=\dfrac{0,3}{0,5}\cdot100\%=60\%\)

b)\(m_{O_2}=0,15\cdot32=4,8g\)

\(\%m_{K_2MnO_4}=\dfrac{29,55}{42,6}\cdot100\%=69,37\%\)

\(\%m_{MnO_2}=100\%-69,37\%=30,63\%\)

b.

4P        + 5O₂ → 2P₂O₅

0,16→    0,2

Dư:      0,025

Sau pứ m_{(bình 1)} = m_P2O5 = 11,36 (g)

O₂        + 2C → 2CO

0,025→ 0,05      0,05

Dư:         0,25

Sau pứ _{m(bình 2}) = m_Cdư = 3 (g)

2KMnO₄ (63/316 mol) \(\underrightarrow{t^o}\) K₂MnO₄ + MnO₂\(\downarrow\) + O₂\(\uparrow\) (63/632 mol).

a. Thể tích khí oxi thu được ở đktc là:

V=63/632.22,4=882/395 (lít).

b. Số mol khí oxi phản ứng là (14,4-11,2)/32=0,1 (mol) > 63/632 (mol).

Kết luận: Giả thiết câu b không xảy ra.