\(n_{CO_2}=\frac{2,016}{22,4}=0,09\left(mol\right)\)

Gọi a,b lần lượt là số mol của Na₂CO₃ và K₂CO₃

Na₂CO₃ + 2HCl -> 2NaCl + CO₂↑ + H₂O

a...............................2a..............a

K₂CO₃ + 2HCl -> 2KCl + CO₂↑ + H₂O

b...............................2b................b

Ta có hệ pt: \(\left\{{}\begin{matrix}106a+138b=10,5\\a+b=0,09\end{matrix}\right.\)

Giải hệ ta được: a = 0,06 ; b = 0,03

=> \(m_{Na_2CO_3}=106.0,06=6,36\left(g\right)\)

\(m_{K_2CO_3}=138.0,03=4,14\left(g\right)\)

=>\(\%m_{Na_2CO_3}=\frac{6,36}{10,5}\cdot100\%=60,57\%\)

\(\%m_{K_2CO_3}=100\%-60,57\%=39,43\%\)

Theo pthh ta có: \(n_{NaCl}=2n_{Na_2CO_3}=2.0,06=0,12mol;n_{KCl}=2n_{K_2CO_3}=2.0,03=0,06mol\)

=>\(m_{NaCl}=0,12.58,5=7,02\left(g\right)\)

\(m_{KCl}=0,06.74,5=4,47\left(g\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

            K₂CO₃ + 2HCl --> 2KCl + CO₂ + H₂O

            CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

Theo PTHH: \(n_{CO_2}=n_{Na_2CO_3}+n_{K_2CO_3}+n_{CaCO_3}=0,05\left(mol\right)\)

=> V = 0,05.22,4 = 1,12 (l)

Mg+2HCl->MgCl2+H2

a..............................a(mol)

Fe+2HCl->FeCl2+H2

b............................b(mol)

=>nCu=3,2/64=0,05mol

=>%mCu=(3,2.100%)/11,2=28,6%

\(=>\left\{{}\begin{matrix}24a+56b=11,2-3,2\\a+b=0,2\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

=>mMg=24.0,1=2,4g=>%Mg=(2,4.100%)/11,2=21,4%

=>%Fe=100%-21,4%-28,6%=50%

b, MgCl2+2NaOH->Mg(OH)2+2NaCL

FeCl2+2NaOH->Fe(OH)2+2NaCl

=>m(kết tủa)=mMg(OH)2+mFe(OH)2

=0,1(58+90)=14,8g

a) mCu= m(k tan)= 3,2(g)

=> m(Mg, Fe)= 11,2- 3,2=8(g)

nH2= 4,48/22,4=0,2(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

a______________2a__a______a(mol)

Fe + 2 HCl -> FeCl2 + H2

b____2b_____b_____b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24a+56b=8\\a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m_{Mg}=24.0,1=2,4\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\end{matrix}\right.\)

=> %mMg= (2,4/11,2).100=21,429%

%mFe= (5,6/11,2).100=50%

=>%mCu= (3,2/11,2).100=28,571%

b/ MgCl2 + 2 NaOH -> Mg(OH)2 + 2 NaCl

0,1___________________0,1(mol)

FeCl2 + 2 NaOH -> Fe(OH)2 +2 NaCl

0,1__________________0,1(mol)

m(kt)=mMg(OH)2 + mFe(OH)2= 58.0,1+ 90.0,1= 14,8(g)