a, PT: \(C+O_2\underrightarrow{t^o}CO_2\)

\(S+O_2\underrightarrow{t^o}SO_2\)

b, Giả sử: \(\left\{{}\begin{matrix}n_C=x\left(mol\right)\\n_S=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow12x+32y=5,6\left(1\right)\)

Ta có: \(n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\)

Theo PT: \(\Sigma n_{O_2}=n_C+n_S=x+y\left(mol\right)\)

\(\Rightarrow x+y=0,3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_C=0,2.12=2,4\left(g\right)\\m_S=0,1.32=3,2\left(g\right)\end{matrix}\right.\)

c, Ta có: \(\left\{{}\begin{matrix}\%m_C=\dfrac{2,4}{5,6}.100\%\approx42,9\%\\\%m_S\approx57,1\%\end{matrix}\right.\)

d, Phần này đề yêu cầu tính theo khối lượng mol hả bạn?

9,2 hay 9,6 oxxi vậy

\(n_{HCl}=0,2.3,5=0,7\left(mol\right)\\ n_{CuO}=a;n_{Fe_2O_3}=b\\ CuO+2HCl\xrightarrow[]{}\Rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\xrightarrow[]{}2FeCl_3+H_2O\\ \left\{{}\begin{matrix}2a+6b=0,7\\80a+160b=20\end{matrix}\right.\\\Rightarrow a=0,05;b=0,1\\ \%_{CuO}=\dfrac{0,05.80}{20}\cdot100=20\%\\ \%_{Fe_2O_3}=\dfrac{0,1.160}{20}\cdot100=80\%\)

a) PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

_____0,02<---0,03<---------------------0,03

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,16}.100\%=25\%\\\%Cu=100\%-25\%=75\%\end{matrix}\right.\)

c) m_H2SO4 = 0,03.98 = 2,94 (g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{2,94}{200}.100\%=1,47\%\)

a, \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

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

b, Gọi số mol KClO₃ và KMnO₄ lần lượt là x,y ( mol ) ( x,y > 0 )

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

x(mol)......................\(\dfrac{3}{2}x\left(mol\right)\)

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

y(mol)..............................................\(\dfrac{1}{2}y\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Tổng số mol O₂ : \(\dfrac{3}{2}x+\dfrac{1}{2}y=0,05\left(1\right)\)

\(m_{KClO_3}=n.M=122,5x\left(g\right)\)

\(m_{KMnO_4}=n.M=158y\left(g\right)\)

\(\Rightarrow122,5x+158y=8,77\left(2\right)\)

Từ (1)(2) ,có :\(\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{2}y=0,05\\122,5x+158y=8,77\end{matrix}\right.\) ( bấm máy tính là ra )

\(\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{KClO_3}=122,5.x=122,5.0,02=2,45\left(g\right)\\m_{KMnO_4}=158y=158.0,04=6,32\left(g\right)\end{matrix}\right.\)

\(\%m_{KClO_3}=\dfrac{2,45}{8,77}.100\%=27\%\)

\(\%m_{KMnO_4}=\dfrac{6,32}{8,77}.100\%=73\%\)

nO2=1,12/22,4=0,05(mol)

2KClO3--->2KCl+3O2

x_______________3/2x

2KMnO4--->K2MnO4+MnO2+O2

y__________________________1/2y

Hệ pt:

\(\left\{{}\begin{matrix}122,5x+158y=8,77\\1,5x+0,5y=0,05\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)

=>mKClO3=0,02.122,5=2,45(g)

=>%mKClO3=2,45/8,77.100%~27,9%

=>%mKMnO4=100%-27,9%=72,1%

a) PTHH: 2KClO₃ --t^o-> 2KCl + 3O₂\(\uparrow\)

................a(mol).........................\(\dfrac{3}{2}a\)(mol)....

2KMnO₄ --t^o--> MnO₂+O₂+K₂MnO₄

....b(mol)......................\(\dfrac{1}{2}b\)(mol)....

b) \(n_{O2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

- Gọi số mol của KClO₃, KMnO₄ lần lượt là a(mol), b(mol)

Từ đó ta có hệ;

\(\left\{{}\begin{matrix}122,5a+158b=8,77\\\dfrac{3}{2}a+\dfrac{1}{2}b=0,05\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,02\left(mol\right)\\b=0,04\left(mol\right)\end{matrix}\right.\)

\(\%KClO_3=\dfrac{0,02.122,5}{8,77}.100\%=27,94\%\)

\(\%KMnO_4=100\%-27,94\%=72,06\%\)

Câu c thì bn dựa vào câu b mà tính nhé

a)

$Cu + 4HNO_3 \to Cu(NO_3)_2 + 2NO_2 + 2H_2O$
$Ag + 2HNO_3 \to AgNO_3 + NO_2 + H_2O$

b)

Gọi $n_{Cu} = a(mol) ; n_{Ag} = b(mol) \Rightarrow 64a + 108b = 4,52(1)$

$n_{NO_2} =2a + b = 0,07(2)$

Từ (1)(2) suy ra a = 0,02 ; b = 0,03

$\%m_{Cu} = \dfrac{0,02.64}{4,52}.100\% = 28,31\%$

$\%m_{Ag} = 71,69\%$

Đáp án A

Cho Mg và Ag tác dụng với HCl chỉ có Mg phản ứng.

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, 24n_Mg + 27n_Al = 6,12 (1)

Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,12\left(mol\right)\\n_{Al}=0,12\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,12.24}{6,12}.100\%\approx47,06\%\\\%m_{Al}\approx52,94\%\end{matrix}\right.\)