mMg = 8,9 . 26,96% = 2,4 (g)

nMg = 2,4/24 = 0,1 (mol)

mZn = 8,9 - 2,4 = 6,5 (g)

nZn = 6,5/65 = 0,1 (mol)

PTHH: 

Mg + H2SO4 -> MgSO4 + H2

Mol: 0,1 ---> 0,1 ---> 0,1 ---> 0,1

Zn + H2SO4 -> ZnSO4 + H2

Mol: 0,1 ---> 0,1 ---> 0,1 ---> 0,1

VH2 = (0,1 + 0,1) . 90% . 22,4 = 4,032 (l)

nH2SO4 = (0,1 + 0,1) . 90% = 0,18 (mol)

VddH2SO4 = 0,18/0,2 = 0,9 (l) = 900 (ml)

mMg = 8,9 . (26,9663/100) = 2,4 g

nMg =2,4/24 = 0,1 mol

n Zn = ( 8,9 -2,4)/65 = 0,1 mol

Mg + H2S04 --> MgS04 + H2

0,1 ---------------------------------> 0,1

Zn + H2S04 ----> ZnS04 + H2 0,1 ----------------------------------> 0,1

VH2 = 22,4 . 0,2 . 0,9 = 4,032 lít

\(n_{H_2SO_4}=n_{H_2}=a(mol)\\ BTKL:\\ m_{hh}+m_{H_2SO_4}=m_Y+m_{H_2}\\ 2,49+98.a= 8,25+2.a\\ \to a=0,06(mol)\\ a/ m_{H_2SO_4}=0,06.98=5,88(g)\\ b/ V_{H_2}=0,06.22,4=1,334(l)\)

a) Ta có : \(m_{KL}+m_{SO^{2-}_4}=m_{muối}\)

=> \(m_{SO_4^{2-}}=8,25-2,49=5,76\left(g\right)\)

=> \(n_{SO_4^{2-}}=\dfrac{5,76}{96}=0,06\left(mol\right)\)

Mặc khác : \(2H^++SO_4^{2-}\rightarrow H_2SO_4\)

=>\(n_{SO_4^{2-}}=n_{H_2SO_4}=0,06\left(mol\right)\)

=> \(m_{H_2SO_4}=0,06.98=5,88\left(g\right)\)

b) Bảo toàn nguyên tố H : \(n_{H_2}=n_{H_2SO_4}=0,06\left(mol\right)\)

=> V_H2 = 0,06.22,4 = 1,344(lít ) 

Đặt \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\Rightarrow65x+24y=8,9\left(1\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ \Rightarrow x+y=0,2\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}65x+24y=8,9\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\\ \Rightarrow\%_{Zn}=\dfrac{0,1\cdot65}{8,9}\cdot100\%\approx73\%\\ \Rightarrow\%_{Mg}=100\%-73\%=27\%\)

\(n_{HCl}=2x+2y=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{14,6\cdot100\%}{14,6\%}=100\left(g\right)\)

\(a)n_{H_2}=\dfrac{7,55}{22,4}=\dfrac{151}{448}mol\\ n_{Mg}=n_{Zn}=a;n_{Fe}=c\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ a.....a\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ a.....a\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b.....b\\ \Rightarrow\left\{{}\begin{matrix}24a+65a+56b=16\\2a+b=\dfrac{151}{448}\end{matrix}\right.\\ \Rightarrow a=0,125;b=\dfrac{39}{448}\\ \%m_{Mg}=\dfrac{24.0,125}{16}\cdot100=18,75\%\\ \%m_{Zn}=\dfrac{65.0,125}{16}\cdot100=50,78\%\\ \%m_{Fe}=100-18,75-50,78=30,47\%\\ b)V_{ddH_2SO_4}=\dfrac{0,125.2+\dfrac{39}{448}}{1}\approx0,337l\)

\(\left\{{}\begin{matrix}Zn\\Fe\\Mg\end{matrix}\right.+H_2SO_4\rightarrow\left\{{}\begin{matrix}ZnSO_4\\FeSO_4\\MgSO_4\end{matrix}\right.+H_2\uparrow\)

Ta có: \(m_{SO_4}=8,25-2,49=5,76\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=n_{H_2}=n_{SO_4}=\dfrac{5,76}{96}=0,06\left(mol\right)\)

a, \(m_{H_2SO_4}=0,06.98=5,88\left(g\right)\)

b, \(V_{H_2}=0,06.22,4=1,344\)

\(n_{Mg}=a;n_{Fe}=0,5a;n_{Zn}=b\\ a\left(24+28\right)+65b=52a+65b=44,2\\ 1,5a+b=\dfrac{24,64}{22,4}1,1\\ a=0,6;b=0,2\\ \%m_{Mg}=\dfrac{24a}{44,2}=32,58\%\\ \%m_{Fe}=\dfrac{28a}{44,2}=38\%\\ \%m_{Zn}=29,42\%\\ m_{ddacid}=\dfrac{98\left(1,5a+b\right)}{0,08}=1347,5g\\ m_{ddsau}=1389,5g\\ C\%_{MgCl_2}=\dfrac{95a}{1389,5}=4,10\%\\ C\%_{FeCl_2}=\dfrac{127.0,5a}{1389,5}=2,74\%\\ C\%_{ZnCl_2}=\dfrac{136b}{1389,5}=1,96\%\)

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