n\(_{H_2SO_4}\) = \(\dfrac{100.29,4\%}{98}\) = 0,3 (mol)

Theo PT: n\(_{H_2}\)= n\(_{H_2SO_4}\)= 0,3 (mol)
\(\Rightarrow\)V\(_{H_2}\)= 0,3.22,4 = 6,72(l)

Theo PT: n\(_{Al}\) = \(\dfrac{2}{3}\)n\(_{H_2}\) = 0,2 (mol)
0,2.27 = 5,4 (g)

m\(_{dd}\) = 5,4+100−0,3.2 = 104,8 (g)

Theo PT: n\(_{Al_2\left(SO_4\right)_3}\) = \(\dfrac{1}{2}\)n\(_{Al}\)

a) pthh: H₂So₄ + Al -> Al₂(so₄)₃ + H₂
m_H2SO4= 29,4%.100: 100%= 29,4(g)
n_H2SO4= 29,4:98=0,3 mol
Ta có: n_H2so4=n_H2= 0,3
=> V_h2= 22,4.0,3= 6,72 l
n_H2SO4= 2/3 n_Al=>n_Al=0,2 mol
=> m_Al=0,2.27=5,4g
b) mdd sau pư= m_Al+ mdd H₂SO₄=5,4+100=105,4 g
C% _H2SO4= 0,3. 98 : 105,4 .100%=30%

a) $Fe + 2HCl \to FeCl_2 + H_2$

b)

n Fe = 8,4/56 = 0,15(mol) ; n HCl = 0,15.2,4 = 0,36(mol)

Ta thấy : 

n Fe / 1 < n HCl /2 nên HCl dư

Theo PTHH : n H2 = n Fe = 0,15 mol

=> V = 0,15.22,4 = 3,36 lít

c) Dung dịch chứa HCl,FeCl2

m dd HCl = D.V = 0,8.150 = 120(gam)

Sau phản ứng : 

n HCl dư = 0,36 - 0,15.2 = 0,06(mol)

n FeCl2 = n Fe = 0,15(mol)

m dd = 8,4 + 120 -0,15.2 = 128,1(gam)

C% HCl = 0,06.36,5/128,1   .100% = 1,71%

C% FeCl2 = 0,15.127/128,1   .100% = 14,87%

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

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot29,4\%}{98}=0,6\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,6}{3}\) \(\Rightarrow\) Axit còn dư, Nhôm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,6-0,3=0,3\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,3\cdot98=29,4\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72 \left(l\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=204,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{204,8}\cdot100\%\approx16,7\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{29,4}{204,8}\cdot100\%\approx14,36\%\end{matrix}\right.\)

a)

$Na_2O + H_2O \to 2NaOH + H_2$
$2Na + 2H_2O \to 2NaOH + H_2$
b)

$n_{Na} = 2n_{H_2} = 0,05.2 = 0,1(mol)$

$\Rightarrow n_{Na_2O} = \dfrac{14,7 - 0,1.23}{62} = 0,2(mol)$
$n_{NaOH} = n_{Na} + 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,2} = 2,5M$

c)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$

$m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$V_{dd\ H_2SO_4}  = \dfrac{122,5}{1,4} = 87,5(ml)$

a) pthh: h2so4+ al = al2(so4)3 + h2
mH2SO4= 29,4%.100: 100%= 29,4(g)
nH2SO4= 29,4:98=0,3 mol
Ta có: nh2so4=nh2=0,3
=> Vh2=22,4.0,3=6,72 l
nH2SO4=2/3nAl=>nAl=0,2 mol
=> mAl=0,2.27=5,4g
b) mdd sau pư= mAl+ mdd H2SO4=5,4+100=105,4 g
C% H2SO4= 0,3. 98 : 105,4 .100%=30%