a,PTHH: C   +  O2  ->  CO2   (*)

            x        x             x    (mol)

           S  +  O2  ->  SO2     (**)

           y       y            y      (mol)

Ta có dB/H2=9 => MB = 29.2=58 (g)=m hh khí B / n hh Khí B

<=>58= \(\frac{44x+64y}{x+y}\)

=>44x+64y=58x+58y

<=>7x=3y

<=> 7x-3y=0   (1)

Mà hh A =12x+32y=13 (2)

b,Từ (1),(2) ta có hệ pt

\(\hept{\begin{cases}7x-3y=0\\12x+3y=13\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,15\\y=0,35\end{cases}}\)

=>%C = (12.0,15)/13 .100%=13,84%

    %S= 100%-13,84%= 86,16%

Từ pt (*),(**) ta có VO2(đktc)=(x+y).22,4=11,2(l)

\(n_{Al}=a\left(mol\right),n_{Ag}=b\left(mol\right)\)

\(m_X=27a+108b=5.4\left(g\right)\left(1\right)\)

\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)

\(a......0.75a...0.5a\)

\(m_{Cr}=0.5\cdot102a+108b=7.5\left(g\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=\dfrac{7}{80},b=\dfrac{9}{320}\)

\(V_{O_2}=0.75\cdot\dfrac{7}{80}\cdot22.4=1.47\left(l\right)\)

\(\%Al=\dfrac{\dfrac{7}{80}\cdot27}{5.4}\cdot100\%=43.75\%\)

\(\%Ag=56.25\%\)

\(n_{FeO\left(bđ\right)}=\dfrac{2.16}{72}=0.03\left(mol\right)\)

\(m_O=2.16-1.84=0.32\left(g\right)\)

\(n_O=\dfrac{0.32}{16}=0.02\left(mol\right)\)

\(n_{H_2}=n_{H_2O}=n_O=0.02\left(mol\right)\)

\(\Rightarrow n_{FeO}=0.02\left(mol\right)\)

\(V_{H_2}=0.02\cdot22.4=0.448\left(l\right)\)

\(H\%=\dfrac{0.02}{0.03}\cdot100\%=66.67\%\)

\(FeO + H_2 \xrightarrow{t^o} Fe + H_2O\\ n_{H_2} = n_{H_2O} = n_{FeO\ pư} = a(mol)\\ \Rightarrow 2a + 2,16 = 1,84 + 18a\\ \Rightarrow a = 0,02(mol)\\ \Rightarrow H = \dfrac{0,02.72}{2,16}.100\% = 66,67\%\ ; V = 0,02.22,4 = 0,448(lít) = 448\ ml\)

\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ Ta\ có :\\ m_O = m_B - m_{hh} = 5,4 - 4,44 = 0,96(mol)\\ n_O = \dfrac{0,96}{32} = 0,03(mol)\\ \Rightarrow n_{Al_2O_3}= \dfrac{1}{3}n_O = 0,01(mol)\\ \Rightarrow n_{Al} = 2n_{Al_2O_3} = 0,02(mol)\\ m_{Al} = 0,02.54 = 1,08(gam)\\ m_{Fe} = 4,44 - 1,08 = 3,36(gam)\)

sai ạ

Al phải là 27 ạ

Gọi CTHH của Z là $C_nH_{2n-2}O_4$ ( Coi là 1 mol)

Ta có :

Bảo toàn e : $n_{O_2} = \dfrac{4n + 2n-2 -2.4}{4} = \dfrac{6n-10}{4}$

Bảo toàn C : $n_{CO_2} = n$

Suy ra: $\dfrac{6n-10}{4} = n \Rightarrow n = 5$

Mặt khác :

Gọi $n_{C_5H_8O_4} = a(mol)$

Suy ra: $n_{CO_2} = 5a ; n_{H_2O} = 4a$
Ta có: $5a.44 + 4a.18 = 43,8 \Rightarrow a = 0,15$

$\Rightarrow m = 0,15.132 = 19,8(gam)$

\(a)n_{KMnO_4} = a; n_{KClO_3} = b\Rightarrow 158a + 122,5b = 99,95(1)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{O_2} = 0,5a +1,5b = \dfrac{14,56}{22,4}=0,65(2)\\ (1)(2)\Rightarrow a = 0,4 ; b = 0,3\\ \%m_{KMnO_4} = \dfrac{0,4.158}{99,95}.100\% = 63,23\%\\ \%m_{KClO_3} = 100\%-63,23\% = 36,77\%\)

\(n_{K_2MnO_4} = n_{MnO_2} = 0,5a = 0,2(mol)\\ n_{KClO_3} = b = 0,3(mol)\\ m_{hh\ sau\ pư} = 99,95 - 0,65.32 = 79,15(gam)\\ \%m_{K_2MnO_4} = \dfrac{0,2.197}{79,15}.100\% = 49,78\%\\ \%m_{MnO_2} = \dfrac{0,2.87}{79,15},100\% = 21,98\%\\ \%m_{KCl} = 28,24\%\)

ko khó lém bn ơi có câu nèo easy hơm ko

\(n_{KClO_3}=\dfrac{245}{122,5}=2mol\)

\(n_{O_2}=\dfrac{53,76}{22,4}=2,4mol\)

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

2                  0           2,4

1,6               1,6        2,4

0,4               1,6         2,4

Thực tế có \(m_{KClO_3}=1,6\cdot122,5=196g\) \(KClO_3\)phản ứng

Hiệu suất phản ứng:

\(H=\dfrac{196}{245}\cdot100\%=80\%\)

\(C1\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,H_2+CuO\rightarrow\left(t^o\right)Cu+H_2O\\ n_{Cu\left(LT\right)}=n_{H_2}=0,3\left(mol\right)\\ n_{Cu\left(TT\right)}=n_{Cu\left(LT\right)}.H=0,3.80\%=0,24\left(mol\right)\\ m_{Cu\left(TT\right)}=0,24.80=19,2\left(g\right)\)

- Oxit:

+ Oxit bazo: Al₂O₃ (Nhôm oxit), CuO (Đồng (II) oxit)

+ Oxit axit: N₂O₅ (dinito pentaoxit)

- Axit: 

H₂CO₃ (Axit cacbonic), H₃PO₄ (axit photphoric)

- Bazo:

KOH (Kali hidroxit), Zn(OH)₂ (Kẽm hidroxit)

- Muối: 

KHSO₄ (Kali hidrosunfat), CuCl₂ (Đồng (II) clorua), ZnSO₄ (Kẽm sunfat)

$n_{O_2} = 0,65(mol)$
Gọi CTPT của Y là $C_nH_{2n-2}O_2$
Bảo toàn electron :

$0,2.(4n + 2n - 2  -2.2) = 4n_{O_2} = 0,65.4 \Rightarrow n = 3,167$

$\to$ Sai đề