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--->K2MnO4 +MnO2 +O2(1)

4O2 + 3Fe----.2Fe3O4(2)

Ta có

n\(_{KMnO4}=\frac{79}{158}=0,2\left(mol\right)\)

Theo pthh

n\(_{o2}=\frac{1}{2}n_{KMnO4}=0,1\left(mol\right)\)

Theo pthh2'

n\(_{Fe}=\frac{3}{4}n_{O2}=0,075\left(mol\right)\)

m\(_{Fe}=0,075.56=4,2\left(g\right)\)

Theo pthh

n\(_{Fe3O4}=n_{O2}=0,05\left(mol\right)\)

m\(_{Fe3O4}=0,05.232=11,6\left(g\right)\)

Chúc bạn học tốt

Ta có : nKMnO4 = 0,5(mol)

PTHH :: 3Fe+2O2--->Fe3O4

KMnO4--->MnO2+K2MnO4+O2

=> nO2 = 0,25(mol)

=> nFe = 0,18(mol)

=>mFe = 10,08(g)

=>nFe2O3 ...

mFe=...

2KClO3---.>2KCl +3O2(1)

3O2+4Al---->2Al2O3(2)

Ta có

n\(_{KClO3}=\frac{49}{122,5}=0,4\left(mol\right)\)

Theo pthh

n\(_{O2}=\frac{3}{2}n_{KCl}=0,6\left(mol\right)\)

Theo pthh2

n\(_{Al2O3}=\frac{2}{3}n_{o2}=0,4\left(mol\right)\)

m\(_{Al2O3}=0,4.102=40,8\left(g\right)\)

Ta có : nKClO3 = 0,6 (mol)

PTHH: 4Al+O2 ---> Al2O3

KClO3---> O2+KCl

=> nO2 = 0,6(mol)

=> nAl2O3 = 0,4(mol)

=>m Al2O3 = 0,4 . 102 = 40,8 (g)

a.KClO3to⟶KCl+1,5O2

BTKL⟶mKClO3=mO2+mA

               =>24,5=mO2+17,3

→O2→KClO3→H=75%

b.

4P        + 5O2 → 2P2O5

0,16→    0,2

Dư:      0,025

Sau pứ m(bình 1) = mP2O5 = 11,36 (g)

O2        + 2C → 2CO

0,025→ 0,05      0,05

Dư:         0,25

Sau pứ m(bình 2) = mCdư = 3 (g)

a, Tính % về khối lượng KCIO₃ đã bị phân hủy ạ

\(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ n_{O_2\left(LT\right)}=\dfrac{3}{4}.0,45=\dfrac{27}{80}\left(mol\right)\\ n_{O_2\left(ban.đầu\right)}=\dfrac{27}{80}.\left(100\%+10\%\right)=\dfrac{297}{800}\left(mol\right)\\ Gọi:n_{KMnO_4}=a\left(mol\right);n_{KClO_3}=b\left(mol\right)\left(a,b>0\right)\\ 2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ \Rightarrow\left\{{}\begin{matrix}158a+122,5b=56,1\\0,8.0,5a+0,85.1,5b=\dfrac{297}{800}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,17087\\b=0,23757\end{matrix}\right.\\ \)

\(n_{Al_2O_3}=\dfrac{0,45}{2}=0,225\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,225=22,95\left(g\right)\\ m_{chất.còn.lại}=m_{Al_2O_3}+m_{KMnO_4\left(còn\right)}+m_{KClO_3\left(còn\right)}\\ \approx22,95+0,2.0,17087.158+0,15.0,23757.122,5\approx32,715\left(g\right)\)

\(2KClO_3\rightarrow3O_2+2KCl\)

\(m_{KClO_3}=m_{O_2}+m_{KCl}\)

\(\Rightarrow m_{KCl}=m_{KClO_3}-m_{KCl}=24,5-9,6=14,9\left(g\right)\)

\(n_{KClO_3}=\dfrac{a}{122,5}mol\)

\(n_{KMnO_4}=\dfrac{b}{158}mol\)

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

\(\dfrac{a}{122,5}\)                     \(\dfrac{3a}{245}\)

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

\(\dfrac{b}{158}\)                                               \(\dfrac{b}{316}\)

Sau phản ứng các chất còn lại bằng nhau.

\(\Rightarrow m_{KCl}=m_{K_2MnO_4}+m_{MnO_2}\)

Theo hai pt: \(\dfrac{a}{122,5}\cdot74,5=\dfrac{b}{158}\cdot\left(197+87\right)\)

\(\Rightarrow\dfrac{a}{b}=1,48\)

\(\dfrac{V_{O_2\left(KMnO_4\right)}}{V_{O_2\left(KClO_3\right)}}=\dfrac{\dfrac{b}{316}}{\dfrac{3a}{245}}=\dfrac{245b}{948a}=\dfrac{1}{1,48}\cdot\dfrac{245}{948}=0,17\)