\(Na_2CO_3+BaCl_2\rightarrow BaCO_3\downarrow+2NaCl\)

\(BaCl_2+CuSO_4\rightarrow BaSO_4\downarrow+CuCl_2\)

\(Na_2CO_3+FeCl_2\rightarrow2NaCl+FeCO_3\downarrow\)

\(MgCl_2+Na_2CO_3\rightarrow MgCO_3\downarrow+2NaCl\)

BaCl₂ + Na₂CO₃ --> BaCO₃ + 2NaCl

CuSO₄ + Na₂CO₃ --> CuCO₃ + Na₂SO₄

2FeCl₃ + 3Na₂CO₃ + 3H₂O --> 2Fe(OH)₃ + 3CO₂ + 6NaCl

2Al(NO₃)₃ + 3Na₂CO₃ + 3H₂O --> 2Al(OH)₃ + 3CO₂ + 6NaNO₃

Na₂CO₃ + MgCl₂ --> MgCO₃ + 2NaCl

BaCl₂ + CuSO₄ --> BaSO₄ + CuCl₂

A

D

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4mol\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

a)Theo pt: \(n_{Fe}=n_{H_2}=0,1mol\)

\(\Rightarrow m_{Fe}=0,1\cdot56=5,6g\)

\(\%m_{Fe}=\dfrac{5,6}{12}\cdot100\%=46,67\%\)

\(\%m_{Cu}=100\%-46,67\%=53,33\%\)

c)\(m_{Cu}=12-5,6=6,4g\Rightarrow n_{Cu}=0,1mol\)

\(BTe:n_{O_2}=n_{Fe}+n_{Cu}=0,2mol\)

\(H=80\%\Rightarrow n_{O_2}=80\%\cdot0,2=0,16mol\)

\(n_{KMnO_4}=\dfrac{1}{2}n_{O_2}=0,08mol\)

\(\Rightarrow m_{KMnO_4}=0,08\cdot158=12,64g\)

cảm ơnnn

B

C

a) 

\(n_{KClO_3}=\dfrac{14,7}{122,5}=0,12\left(mol\right)\)

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

             0,12---------------->0,18

=> V_O2 = 0,18.22,4 = 4,032 (l)

b) 

PTHH: 4R + nO₂ --t^o--> 2R₂O_n

       \(\dfrac{0,72}{n}\)<--0,18

=> \(M_R=\dfrac{14,4}{\dfrac{0,72}{n}}=20n\left(g/mol\right)\)

Chọn n = 2 => M_R= 40 (g/mol)

=> R là Ca

c) 

Gọi số mol CH₄, H₂ là a, b (mol)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

              a--->2a

             2H₂ + O₂ --t^o--> 2H₂O

              b-->0,5b

=> \(\left\{{}\begin{matrix}2a+0,5b=0,18\\16a+2b=1,12\end{matrix}\right.\)

=> a = 0,05 (mol); b = 0,16 (mol)

\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,05+0,16}.100\%=23,81\%\\\%V_{H_2}=\dfrac{0,16}{0,05+0,16}.100\%=76,19\%\end{matrix}\right.\)

cảm ơnnn

B

nhh khí = 8,96/22,4 = 0,4 (mol)

nCaCO3 = 20/100 = 0,2 (mol)

Ca(OH)2 + CO2 -> CaCO3 + H2O

0,2 <--- 0,2 <--- 0,2

2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O

0,1 <--- 0,25 <--- 0,2

nH2 = 0,4 - 0,1 = 0,3 (mol)

2H2 + O2 -> (t°) 2H2O

0,3 ---> 0,15

nO2 = 0,15 + 0,25 = 0,4 (mol)

Vkk = 0,4 . 5 . 22,4 = 44,8 (l)

Chọn C

C

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

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

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(\Rightarrow\left\{{}\begin{matrix}27x+56y=22,2\\\dfrac{1}{2}x\cdot102+\dfrac{1}{3}y\cdot232=33,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

a)\(\%m_{Al}=\dfrac{0,2\cdot27}{22,2}\cdot100\%=24,32\%\)

\(\%m_{Fe}=100\%-24,32\%=75,68\%\)

b)Theo hai pt trên:

\(\Rightarrow n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{2}{3}n_{Fe}=\dfrac{3}{4}\cdot0,2+\dfrac{2}{3}\cdot0,3=0,35mol\)

\(H=80\%\Rightarrow n_{O_2}=80\%\cdot0,35=0,28mol\)

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

\(\dfrac{14}{75}\)                         0,28

\(m_{KClO_3}=\dfrac{14}{75}\cdot122,5=22,87g\)

232 với 102 đâu ra v ạ

nKMnO4 = 31,6/158 = 0,2 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nO2 (LT) = 0,2/2 = 0,1 (mol)

nO2 (TT) = 0,1 . 80% = 0,08 (mol)

mO2 (TT) = 0,08 . 32 = 2,56 (g)

\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2mol\)

\(H=80\%\Rightarrow n_{KMnO_4}=0,2\cdot80\%=0,16mol\)

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

0,16                                               0,08

\(m_{O_2}=0,08\cdot32=2,58g\)