nH2SO4=0,4*1=0,4 mol

nH2=6,72/22,4=0,3 mol

2Al + 3H2SO4 -->Al2(So4)3 + 3H2

 0,2                                             0,3          mol 

=> mAl = 0,2*27=5,4 g

=> mCu =5,9-5,4=0,5 g

BaCl2 + H2SO4 --> BaSO4 + 2HCl

                0,4              0 ,4                       mol

=> m BaSO4 = 0,4 * 233=93,2 g      

\(a) 2Al+ 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)\\ m_{Al} = 0,2.27 = 5,4(gam)\\ m_{Cu} = 5,9 - 5,4 = 0,5(gam)\\ b) \)

Bảo toàn nguyên tố với S :

\(n_{BaSO_4} = n_{H_2SO_4} = 0,4(mol)\\ m = 0,4.233 = 93,2(gam)\)

\(a) 2Al+ 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ n_{Al} = \dfrac{2}{3}n_{H_2} = 0,2(mol)\\ m_{Al} = 0,2.27 = 5,4(gam)\\ m_{Cu} = 5,9 - 5,4 = 0,5(gam)\\ b) \)

Bảo toàn nguyên tố với S :

\(n_{BaSO_4} = n_{H_2SO_4} = 0,4(mol)\\ m = 0,4.233 = 93,2(gam)\)

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

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

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, \(V_{H_2}=0,06.22,4=1,344\)

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

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 ) 

Bài 2 : 

a)

$Mg + 2HCl \to MgCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{HCl} = 0,5.36,5 = 18,25(gam)$

b)

Bảo toàn khối lượng : 

$m_A = 22,85 + 0,25.2 - 18,25 = 5,1(gam)$