Ta có: \(n_{Al\left(OH\right)_3}=\dfrac{15,6}{78}=0,2\left(mol\right)\), \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

BTNT Al, có: n_Al = n_Al(OH)3 = 0,2 (mol)

BT e,có: n_Na + 3n_Al = 2n_H2 ⇒ n_Na = 0,2 (mol)

\(\Rightarrow m=m_{Na}+m_{Al}=10\left(g\right)\)

nH2=8,9622,4=0,4(mol)��2=8,9622,4=0,4(���)

BTNT Al, có: n_Al = n_Al(OH)3 = 0,2 (mol)

BT e,có: n_Na + 3n_Al = 2n_H2 ⇒ n_Na = 0,2 (mol)

⇒m=mNa+mAl=10(g)

Đặt nMg=a(mol); nAl=b(mol)

PTHH: Mg +2 HCl -> MgCl2 + H2

a________2a_______a_____a(mol)

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

b_____3b____b_____1,5b(mol)

Ta có hpt: \(\left\{{}\begin{matrix}24a+27b=7,8\\a+1,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

=> %mMg=[(0,1.24)/7,8].100=30,769%

=>%mAl= 69,231%

c) MgCl2 + 2 NaOH -> Mg(OH)2 + 2 NaCl

0,1_______________0,1(mol)

AlCl3 + 3 NaOH -> Al(OH)3 + 3 NaCl

0,2____________0,2(mol)

=> m=m(kết tủa)= mMg(OH)2+ mAl(OH)3= 58.0,1+ 78.0,2= 21,4(g)

Đáp án B

Rắn X phải có A1 dư vì chỉ có A1 mới phản ứng với dung dịch NaOH tạo khí H₂. Mặt khác do phản ứng hoàn toàn nên Fe₃O₄ hết.

Gọi $n_{Na} = a(mol)$

2Na + 2H₂O → 2NaOH + H₂

a...........................a..........0,5a.....(mol)

2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

..a...........a............................................1,5a....(mol)

Suy ra : $0,5a + 1,5a = \dfrac{3,36}{22,4} = 0,15 \Rightarrow a = 0,075$

Vậy :

$m = 0,075.23 + 0,075.27 + 1,35 = 5,1(gam)$

Gọi nNa=a(mol)���=�(���)

2Na + 2H₂O → 2NaOH + H₂

a...........................a..........0,5a.....(mol)

2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

..a...........a............................................1,5a....(mol)

Suy ra : 

PTHH: \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\uparrow\)  (1)

            \(Na_2SO_4+Ba\left(OH\right)_2\rightarrow2NaOH+BaSO_4\downarrow\)  (2)

            \(2NaOH+MgCl_2\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)  (3)

            \(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)  (4)

Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}\cdot\dfrac{2,3}{23}=0,05\left(mol\right)\\n_{BaCl_2}=\dfrac{60\cdot14,25\%}{208}=0,05\left(mol\right)\\n_{MgCl_2}=\dfrac{30\cdot19\%}{95}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) PT (2) p/ứ hết; PT (3) có MgCl₂ dư 0,01 mol  

\(\Rightarrow n_{MgO}=n_{Mg\left(OH\right)_2}=n_{BaSO_4}=0,05\left(mol\right)\)

\(\Rightarrow m_{rắn}=m_{MgO}+m_{BaSO_4}=0,05\cdot\left(40+233\right)=13,65\left(g\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{NaCl}=n_{Na}=0,1\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,05\left(mol\right)=n_{H_2SO_4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,1\cdot58,5=5,85\left(g\right)\\m_{ddH_2SO_4}=\dfrac{0,05\cdot98}{4,9\%}=100\left(g\right)\\m_{Mg\left(OH\right)_2}=0,05\cdot58=2,9\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na}+m_{ddH_2SO_4}+m_{ddBaCl_2}+m_{ddMgCl_2}-m_{BaSO_4}-m_{Mg\left(OH\right)_2}=177,75\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{5,85}{177,75}\cdot100\%\approx3,29\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{0,01\cdot95}{177,75}\cdot100\%\approx0,53\%\end{matrix}\right.\)