$\rm a)$
$\rm 2Mg + CO_2 \xrightarrow{t^o} 2MgO + C$
$\rm C + CO_2 \xrightarrow{t^o} 2CO$
$\rm CO + CuO \xrightarrow{t^o} Cu + CO_2$
$\rm CO_2 + Ca(OH)_2 \rightarrow CaCO_3 \downarrow + H_2O$
$\rm CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2$
$\rm Ca(HCO_3)_2 \xrightarrow{t^o} CaCO_3 \downarrow + CO_2 \uparrow + H_2O$
$\rm b)$
$\rm CO_2 + H_2O + Na_2CO_3 \rightarrow 2NaHCO_3$
$\rm NaHCO_3 + NaOH \rightarrow Na_2CO_3 + H_2O$
$\rm Na_2CO_3 + Ca(HCO_3)_2 \rightarrow CaCO_3 \downarrow + 2NaHCO_3$
$\rm NaHCO_3 + HCl \rightarrow NaCl + CO_2 \uparrow + H_2O$

\(PTHH:2Al+H_2SO_4-^{t^o}>Al_2SO_4+H_2\uparrow\)

ap dung DLBTKL ta co

\(m_{Al}+m_{H_2SO_4}=m_{Al_2SO_4}+m_{H_2}\)

thay so ta co

\(5,4+m_{H_2SO_4}=34,2+0,6\\ =>m_{H_2SO_4}=34,2+0,6-5,4\\ =>m_{H_2SO_4}=29,4\left(g\right)\)

Ta có: \(\left\{{}\begin{matrix}m_{bình.đựng.H_2SO_{4\left(đặc\right)}.tăng}=m_{H_2O}=2,52\left(g\right)\\m_{CaCO_3}=m_{kết.tủa}=14\left(g\right)\end{matrix}\right.\)

Theo BTNT: \(\left\{{}\begin{matrix}n_H=2n_{H_2O}=2.\dfrac{2,52}{18}=0,28\left(mol\right)\\n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{14}{100}=0,14\left(mol\right)\\n_{O\left(X\right)}=\dfrac{3,0,8-0,28-0,14.12}{16}=0,07\left(mol\right)\end{matrix}\right.\)

=> \(n_C:n_H:n_O=0,14:0,28:0,07=2:4:1\)

=> CTPT của X có dạng \(\left(C_2H_4O\right)_n\)

Mặt khác: \(n_X=n_{N_2}=\dfrac{1,54}{28}=0,055\left(mol\right)\)

=> \(M_X=\dfrac{4,84}{0,055}=88\left(g/mol\right)\)

=> \(n=\dfrac{88}{44}=2\left(t/m\right)\)

=> CTPT của X là \(C_4H_8O_2\)

M(X)= 0,552 x 29= 16 (g/mol)

a) \(n_{Al}=\dfrac{m}{M}=\dfrac{33,75}{27}=1,25\left(mol\right)\)

PTHH: \(4Al+3O_2\rightarrow2Al_2O_3\)

Theo PTHH: \(n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3}{4}.1,25=0,9375\left(mol\right)\)

=> \(V_{O_2\left(đktc\right)}=n.22,4=0,9375.22,4=21\left(l\right)\)

Mà \(V_{O_2}=\dfrac{1}{5}V_{kk}\)

=> \(V_{kk}=5.V_{O_2}=5.21=84\left(l\right)\)

b) Theo PTHH: \(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.1,25=0,625\left(mol\right)\)

=> \(m_{Al_2O_3}=n.M=0,625.102=63,75\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{O_2}=a\left(mol\right)\\n_{N_2}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)

=> \(\left\{{}\begin{matrix}m_{O_2}=32a\left(g\right)\\m_{N_2}=28b\left(g\right)\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}n_{hh.khí}=n_{O_2}+n_{N_2}=a+b=0,8\\m_{hh.khí}=m_{O_2}+m_{N_2}=32a+28b=23,2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,6\left(mol\right)\end{matrix}\right.\left(TM\right)\)

=> \(\left\{{}\begin{matrix}m_{O_2}=32a=32.0,2=6,4\left(g\right)\\m_{N_2}=m_{hh.khí}-m_{O_2}=23,2-6,4=16,8\left(g\right)\end{matrix}\right.\)

Mai kiểm tra 15' rồi hy vọng mn giúp

EEm cần bài nào thế em?

5.1 với 5.2 ạ

Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\Rightarrow27a+56b=13,7\left(1\right)\)

\(n_{H_2}=\dfrac{11,1555}{24,79}=0,45\left(mol\right)\)

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

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

Theo PTHH: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}\Rightarrow1,5a+b=0,45\left(2\right)\)

Từ (1), (2) ta có hệ PT: \(\left\{{}\begin{matrix}27a+56b=13,7\\1,5a+b=0,45\end{matrix}\right.\)

Giải hệ PT, ta được: \(\left\{{}\begin{matrix}a=\dfrac{23}{114}\left(mol\right)\\b=\dfrac{14}{95}\left(mol\right)\end{matrix}\right.\left(TM\right)\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{\dfrac{23}{114}.27}{13,7}.100\%=39,76\%\\\%m_{Fe}=100\%-39,76\%=60,24\%\end{matrix}\right.\)