1)

$n_{H_2SO_4} = n_{H_2} = 0,45(mol)$

$\Rightarrow n_{BaSO_4} =n_{Ba(OH)_2} = n_{H_2SO_4} = 0,45(mol)$

Gọi $n_{Al} = 2a(mol) \to n_{Fe} =a (mol)$

Kết tủa F gồm : 

$Fe(OH)_2 : a(mol)$
$Al(OH)_3 : b(mol) \Rightarrow n_{Ba(AlO_2)_2} = \dfrac{2a-b}{2}(mol)$

Nung F, thu được chất rắn gồm : 

$Fe_2O_3 : 0,5a(mol)$
$Al_2O_3 : 0,5b(mol)$
$BaSO_4:  0,45(mol)$
$\Rightarrow 0,5a.160 + 0,5b.102 + 0,45.233 = 117,95(1)$
$n_{Ba(OH)_2} =n_{FeSO_4} + 3n_{Al_2(SO_4)_3} + n_{Ba(AlO_2)_2} = a + 3a + \dfrac{2a - b}{2} = 0,45(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,1

$0,5n_A + 1,5n_{Al} + n_{Fe} = n_{H_2}$

$\Rightarrow n_A = 0,1(mol)$

Ta có : 0,1.A + 0,1.2.27 + 0,1.56 = 13,3

Suy ra: A = 23(Natri)

2)

$\%m_{Na} = \dfrac{0,1.23}{13,3}.100\% = 17,3\%$
$\%m_{Al} = \dfrac{0,2.27}{13,3}.100\% = 40,6\%$

$\%m_{Fe} = 100\% - 17,3\ - 40,6\% = 42,4\%$

2M+H2SO4\(\rightarrow\)M2SO4+H2(1)

u ___________0,5u____0,5u

2Al+3H2SO4\(\rightarrow\)Al2(SO4)3+3H2(2)

v _____________0,5v______1,5v

nH2=\(\frac{2,464}{22,4}\)=0,11\(\rightarrow\)0,5u=1,5v=0,11(II)

ddY chứa 0,5u mol M2SO4 và 0,5v mol Al2(SO4)3

M2SO4+Ba(OH)2\(\rightarrow\)2MOH+BaSO4

0,5u________________ u___ 0,5u

Al2(SO4)3+3Ba(OH)2\(\rightarrow\)2Al(OH)3+3BaSO4

0,5v _________________v_________1,5v

MOH+Al(OH)3\(\rightarrow\)MAlO2+2H2O

u ______u

\(\rightarrow\)nAl(OH)3 dư=v-u(mol)

nBaSO4=0,5u+1,5v=0,11

mBaSO4=0,11.233=25,63(g)

mAl(OH)3 dư=27,19-25,63=1,56g

nAl(OH)3 dư=0,02\(\rightarrow\)v-u=0,02(III)

(II)(III)\(\rightarrow\)u=0,04 v=0,06

Thay u v vào (I)

0,04M+0,06.27=3,18

\(\rightarrow\)M=39

\(\rightarrow\)M là Kali(K)

a, Ta có: 27n_Al + 56n_Fe = 0,83 (1)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Fe}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow n_{Al}=n_{Fe}=0,01\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%\approx32,53\%\\\%m_{Fe}\approx67,47\%\end{matrix}\right.\)

b, n_H2SO4 = n_H2 = 0,025 (mol)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,025.98}{20\%}=12,25\left(g\right)\)

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

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 )