nO2 = 6.72/22.4 = 0.3 (mol) 

BTKL : 

mKMnO4 = 116.8 + 0.3*32 = 126.4 (g) 

nKMnO4 = 126.4/158 = 0.8 (mol) 

2KMnO4 -to-> K2MnO4 + MnO2 + O2 

0.6_________________________0.3 

H% = 0.6/0.8 * 100% = 75% 

\(n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol) \\m = m_{chất\ rắn} + m_{O_2} = 116,8 + 0,3.32 = 126,4(gam)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,3.2 = 0,6(mol)\\ H = \dfrac{0,6.158}{126,4}.100\%= 75\%\)

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 

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

\(\rightarrow a=0,15\)

\(b=0,16\)

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

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

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

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

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

Gọi số mol KMnO4 phản ứng là a (mol).

\(2KMnO_4\underrightarrow{t^{^0}}K_2MnO_4+MnO_2+O_2\)

\(n_{KMnO_4bđ}=\dfrac{79}{158}=0,5mol\\ n_{K_2MnO_4}=n_{MnO_2}=n_{O_2}=a\left(mol\right)\\ m_{rắn}=m_{K_2MnO_4}+m_{MnO_2}+m_{KMnO_4dư}=197.0,5a+87.0,5a+158\left(0,5-a\right)=74,2\\ a=0,3\left(mol\right)\\ H=\dfrac{0,3}{0,5}.100\%=60\%\)

Đáp án là D. 50%.

Gọi x, y là số mol lần lượt của KMnO4 và KClO3

\(158x+122,5y=43,3\left(1\right)\)

\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

0,9x_____________________0,45x____
\(2KClO_3\rightarrow2KCl+3O_2\)

0,9y_______________1,35y

\(\%_{Mn}=\frac{55a}{43,3-32\left(0,45x+1,35y\right)}=24,103\left(2\right)\)

Giải hệ phương trình : (1) + (2)

\(\Rightarrow\left\{{}\begin{matrix}x=0,15\\b=0,16\end{matrix}\right.\)

\(\Rightarrow\%_{KMnO4}=54,73\%\Rightarrow\%_{KClO3}=45,27\%\)

Đặt \(n_{Fe}=x\left(mol\right)\)

Rắn gồm \(\left\{{}\begin{matrix}Fe\\Fe_2O_3\left(dư\right)\end{matrix}\right.\)

\(PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ \left(mol\right)....0,5x.........\leftarrow x\)

\(m_{Fe}+m_{Fe_2O_3\left(dư\right)}=m_{rắn}\\ \Leftrightarrow56x+\left(24-0,5x.160\right)=19,2\\ \Leftrightarrow56x+24-80x=19,2\\ \Leftrightarrow24x=4,8\\ \Leftrightarrow x=0,2\)

\(H=\dfrac{m_{Fe_2O_3\left(pư\right)}}{m_{Fe_2O_3}}.100\%=\dfrac{m_{Fe_2O_3}-m_{Fe_2O_3\left(dư\right)}}{m_{Fe_2O_3}}.100\%=\left(1-\dfrac{m_{Fe_2O_3\left(dư\right)}}{m_{Fe_2O_3}}\right).100\%=\left(1-\dfrac{24-0,5.0,2.160}{24}\right).100\%=\dfrac{200}{3}\approx66,67\%\)

mCaCO3 = 500*80%= 400 (g) 

nCaCO3 = 400/100 = 4 (mol) 

nCaCO3(pư) = 4*70%=2.8 (mol) 

CaCO3 -to-> CaO + CO2 

2.8..................2.8 

Chất rắn X : CaCO3 dư , CaO 

mX = ( 4 -2.8 ) *100 + 2.8*56 = 276.8 (g) 

%CaO = 2.8*56/276.8 * 100% = 56.64%

a)mCaCO3=500.80%=400(g) -> nCaCO3=400/100=4(mol)

PTHH: CaCO3 -to-> CaO + H2O

nCaO(LT)=nCaCO3=4(mol)

=> nCaO(TT)=4. 70%=2,8(mol)

=>mX=mCaO+ m(trơ)+ mCaCO3(chưa p.ứ)=2,8.56+100+ 1,2.100=376,8(g)

b) %mCaO= (156,8/376,8).100=41,614%