Ban đầu:\(n_{H+}=n_{HCL}+2nH_2SO_4=0,25.\left(0,08+2.0,01\right)=0,025\left(mol\right)\)

\(n_{OH-}=2n_{Ba\left(OH\right)_2}=0,25a.2=0,5a\)

\(n_{Ba^2+}=n_{Ba\left(OH\right)_2}=0,25a;nSO^2_4=nH_2SO_4=0,0025\left(mol\right)\)

Dung dịch sau phản ứng có pH = 12 => pOH = 2

\(\Rightarrow\left[OH^-\right]=10^{-2}=0,01\left(M\right)\Rightarrow n_{OH-}=0,01.0,5=0,005\left(mol\right)\)Vì pH = 12 > 7 nên \(H^+hết,OH^-còn\)

\(H^++OH^-\rightarrow H_2O\)

0,025-->0,025

\(\Rightarrow n_{OH^-}còn=0,5a-0,025=0,005\Rightarrow a=0,06\left(mol/l\right)\)

Từ đó suy ra được \(nBa^{2+}=0,25.0,06=0,015\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

0,0025<--0,0025

\(\Rightarrow m=m_{Baso_4}=0,025.233=0,5825\left(gam\right)\)

Ban đầu:\(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,25.\left(0,08+2.0,01\right)=0,025\left(mol\right)\)

\(n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,25a.2=0,5a\)

\(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,25a\); \(n_{SO_4^{2-}}=n_{H_2SO_4}=0,0025\left(mol\right)\)

Dung dịch sau phản ứng có pH = 12 => pOH = 2

\(\Rightarrow\left[OH^-\right]=10^{-2}=0,01\left(M\right)\Rightarrow n_{OH^-}=0,01.0,5=0,005\left(mol\right)\)Vì pH = 12 > 7 nên \(H^+\) hết, \(OH^-\) còn.

\(H^++OH^-\rightarrow H_2O\)

0,025-->0,025

=> \(n_{OH^-}\text{còn}=0,5a-0,025=0,005\Rightarrow a=0,06\left(mol\text{/}l\right)\)

Từ đó suy ra được \(n_{Ba^{2+}}=0,25.0,06=0,015\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

0,0025<--0,0025

=> \(m=m_{BaSO_4}=0,0025.233=0,5825\left(gam\right)\)

nBa = 0,1 (mol)

Ba + 2H2O ---> Ba(OH)2 + H2 (1)

0,1........................0,1

nAl2(SO4)3 = 0,04 (mol)

3Ba(OH)2 + Al2(SO4)3 ---> 2Al(OH)3 + 3BaSO4

0,1...............................................\(\dfrac{1}{15}\).............0,1

Kết tủa là Al(OH)3 và BaSO4

\(\Rightarrow m=\dfrac{1}{15}.78+0,1.233=28,5\left(g\right)\)

+) Ta có: \(V_{ddAl_2\left(SO_4\right)_3}=200\left(ml\right)=0,2\left(l\right)\\ =>n_{Al_2\left(SO_4\right)_3}=0,2.0,2=0,04\left(mol\right)\\ n_{Ba}=\dfrac{13,7}{137}=0,1\left(mol\right)\\ PTHH:3Ba+Al_2\left(SO_4\right)_3->3BaSO_4\downarrow+2Al\\ \dfrac{0,1}{3}< \dfrac{0,04}{1}\)

=> Al₂(SO₄)₃ dư, Ba hết tính theo n_Ba.

Ta có: \(n_{BaSO_4}=n_{Ba}=0,1\left(mol\right)\\ =>m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

\(n_{CuCl_2}=\dfrac{100.2}{1000}=0,2\left(mol\right)\)

a, PTHH:

CuCl₂ + 2NaOH \(\rightarrow\) Cu(OH)₂↓ + 2NaCl

0,2..............0,4.............0,2

b, m_{dd NaOH} = 0,4.40 . \(\dfrac{100}{20}\) =80 (g)

c, Kết tủa A là Cu(OH)₂

PT:

Cu(OH)₂\(\underrightarrow{t^o}\) CuO + H₂O

0,2.................0,2

m_CuO = 0,2 . 80 = 16 (g)

\(a)n_{Ba\left(OH\right)_2}=0,05\cdot0,2\cdot2=0,02mol\\ pH=1\Rightarrow\left[OH^-\right]=0,1M\Rightarrow n_{HCl}=0,1\cdot0,3=0,03mol\\ n_{Ba\left(OH\right)_2}+n_{HCl}=0,02+0,03=0,05mol\\ \Rightarrow C_M=\dfrac{0,05}{0,5}=0,1M\Rightarrow pH=1\)

Bạn xem lại đề !

PTHH

Al + 6HCl -> 2AlCl₃ + 3H₂

Mg + 2HCl -> MgCl₂ + H₂

n_H2 = 11.2/22.4 = 0.5 mol

gọi nAl = x mol ; nMg = y mol

ta có 27x + 24y = 10.2

3x+ y = 0.5

-> x = 0.04 ; y = 0.38

mAl = 0.04.27 =1.08 (g)

%mAl = (1.08/10.2).100%=10.6%

%mMg = 100% - 10.6% = 89.4%

Thiếu :Hãy tính tp %theo kl cua mõi kim loai (.) hh