gọi số mol của Mg là a mol , Zn là b  mol

=> 24a + 65b=21,4 

nH2=1,1/2=0,55 

Mg + H2SO4 --> MgSO4 +H2

a                                         a         mol

Zn + H2SO4 --> ZnSO4 +H2

b                                         b          mol

=> a + b = 0,55 

=> a=0,35  mol ,b=0,2 mol

=> mMg = 0,35 *24=8,4 g

mZn =0,2 * 65= 13 g

mMgSO4 = 0,35 * 120=42

mZnSO4=0,2*161=32,2

=>m muối = 42 + 32,2=74,2 g

n H2SO4 = 0,35 + 0,2=0,55 mol

=>VH2SO4 = 0,55 *22,4=12,32 => V H2SO4 thực =12,32+ 12,32*10%=14,652 g

câu 2

n khí =3,36/22,4=0,15 mol

  Zn +  H2SO4   -->     ZnSO4 + H2  (1)

0,15                              0,15       0,15         mol

=>mZn = 0,15 *65=9,75 g

=> mCuO=15,75-9,75=6g => nCuO=6/80=0,075 mol

CuO + H2SO4 --> CuSO4 +H2O

0,075                         0,075              mol

=>mCuSO4=0,075*160=12g

(1)=> mZnSO4 = 0,15*161=24,15 g

=> m muối =12 + 24,15=36,15 g

mình sửa lại phần cuối câu 1 nhá

\(\Sigma\)nH2SO4=0,35+0,2=0,55 mol

VH2SO4=0,55/0,01=55 lít

v H2SO4 thực tế = 55+ 55*10%=60,5 lít

gfvfvfvfvfvfvfv555

\(Đặt:n_{Fe}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Kim.loại.còn.lại.sau.p.ứ:Cu\\ n_{Cu}=\dfrac{25,4}{64}=0,4\left(mol\right)\\ a,PTHH:Fe+CuSO_4\rightarrow FeSO_4+Cu\\ 2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\\ \Rightarrow\left\{{}\begin{matrix}56a+27b=11\\a+1,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\b, \%m_{Fe}=\dfrac{0,1.56}{11}.100\approx50,909\%\\ \%m_{Cu}\approx100\%-50,909\%\approx49,091\%\\ c,V_{ddsau}=V_{ddCuSO_4}=0,2\left(l\right)\\ C_{MddFeSO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right);C_{MddAl_2\left(SO_4\right)_3}=\dfrac{0,2:2}{0,2}=0,5\left(M\right)\\ d,C_{MddCuSO_4}=\dfrac{a+1,5b}{0,2}=2\left(M\right)\)

1,7 gam chất không tan là Cu

=> \(\%m_{Cu}=\dfrac{1,7}{10}.100=17\%\)

Fe+ 2HCl ---------> FeCl2 + H2 ;

2Al + 6HCl ---------> 2AlCl3 + 3H2

Gọi x,y lần lượt là số mol Fe, Al

\(\left\{{}\begin{matrix}56x+27y=10-1,7\\x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\%m_{Fe}=\dfrac{0,1.56}{10}.100=56\%\)

\(\%m_{Al}=\dfrac{0,1.27}{10}.100=27\%\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\n_{H_2}=\dfrac{2,16}{22,4}=0,09\left(mol\right)\\ \Rightarrow \left\{{}\begin{matrix}1,5a+b=0,09\\27a+56b=2,76\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,03\end{matrix}\right.\\ \Rightarrow\%m_{Al}=\dfrac{0,04.27}{2,76}.100\approx39,13\%\\ \Rightarrow\%m_{Fe}\approx100\%-39,13\%\approx60,87\%\)

\(b,V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ n_{AlCl_3}=n_{Al}=0,04\left(mol\right);n_{FeCl_2}=n_{Fe}=0,03\left(mol\right)\\ \Rightarrow C_{MddFeCl_2}=\dfrac{0,03}{0,2}=0,15\left(M\right)\\ C_{MddAlCl_3}=\dfrac{0,04}{0,2}=0,2\left(M\right)\)

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

  \(x\)                                      \(1,5x\)

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

 \(y\)                                    \(y\)

Có \(27x+56y=2,76\left(1\right)\)

\(1,5x+y=\dfrac{2,016}{22,4}=0,09\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,03\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,04\cdot27}{2,76}\cdot100\%=39,13\%\)

\(\%m_{Fe}=100\%-39,13\%=60,87\%\)

2Al+6HCl->2AlCl₃+3H₂

x--------------------------3\2 x mol

Fe+2HCl->FeCl₂+H₂

y-------------------------y mol

n _H2=\(\dfrac{2,016}{22,4}\)=0,09 mol

=>\(\left\{{}\begin{matrix}27x+56y=2,76\\\dfrac{3}{2}x+y=0,09\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0,04\\y=0,03\end{matrix}\right.\)

=>% m _Al =\(\dfrac{0,04.27}{2,76}100\)=39,13%

=>% m _Fe=60,87%

=>Cm AlCl3 =\(\dfrac{0,04}{0,2}\)=0,2M

=>CM _Fecl2=\(\dfrac{0,03}{0,2}\)=0,15M

\(n_{NaCl}=a\left(mol\right)\)

\(n_{KCl}=b\left(mol\right)\)

\(m_{hh}=58.5a+74.5b=30.325\left(g\right)\left(1\right)\)

\(n_{AgCl}=\dfrac{64.575}{143.5}=0.45\left(mol\right)\)

\(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\)

\(KCl+AgNO_3\rightarrow KNO_3+AgCl\)

\(\Rightarrow a+b=0.45\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.25\)

\(m_{NaCl}=0.2\cdot58.5=11.7\left(g\right)\)

\(m_{KCl}=0.25\cdot74.5=18.625\left(g\right)\)

Gọi x, y lần lượt là số mol NaCl, KCl

\(NaCl+AgNO_3\rightarrow AgCl+NaNO_3\)

\(KCl+AgNO_3\rightarrow AgCl+KNO_3\)

\(n_{AgCl}=\dfrac{64,575}{143,5}=0,45\left(mol\right)\)

Ta có hệ \(\left\{{}\begin{matrix}58,5x+74,5y=30,325\\x+y=0,45\end{matrix}\right.\)

=> x=0,2 ; y=0,25

=> \(m_{NaCl}=11,7\left(g\right);m_{KCl}=18,625\left(g\right)\)

\(\left\{{}\begin{matrix}n_{NaCl}=a\left(mol\right)\\n_{KCl}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)

PTHH:

\(NaCl+AgNO_3\rightarrow NaNO_3+AgCl\\ ...a.....a.........a.....a\left(mol\right)\\ KCl+AgNO_3\rightarrow KNO_3+AgCl\\ b.........b........b..............b\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}58,5a+74,5b=30,325\\143,5a+143,5b=64,575\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,25\end{matrix}\right.\\ \left\{{}\begin{matrix}m_{NaCl}=58,5.0,2=11,7\left(g\right)\\m_{KCl}=74,5.0,25=18,625\left(g\right)\end{matrix}\right.\)

\(n_{CuSO_4}=x\left(mol\right)\)

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

\(x.....x...........x........x\)

\(m_{tăng}=m_{Cu}-m_{Fe}=10.24-10=0.24\left(g\right)\)

\(\Leftrightarrow64x-56x=0.24\)

\(\Leftrightarrow x=0.03\)

\(m_{Fe\left(pư\right)}=0.03\cdot56=1.68\left(g\right)\)

\(m_{Cu}=0.03\cdot64=1.92\left(g\right)\)

\(m_{CuSO_4}=0.03\cdot160=4.8\left(g\right)\)

\(m_{dd_{CuSO_4}}=\dfrac{4.8\cdot100}{5}=96\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=1.68+96-1.92=95.76\left(g\right)\)

\(C\%_{FeSO_4}=\dfrac{0.03\cdot152}{95.76}\cdot100\%=4.76\%\)