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

Ba(OH)2+2Al(OH)3->Ba(AlO2)2+4H2O

Số mol BaSO4=69,9/233=0,3mol

=> số mol Al(OH)3= 0,2mol

Tổng số mol Ba(OH)2=n(BaSO4)+0,5n(Ba(OH)2)= 0,3+0,1=0,4mol

V=n/c=0,4/0,4 =1lit

Câu 3 : 

\(pH=-log\left[H^+\right]=-log\left(0.1\right)=1\)

Câu 4 : 

Chứa các ion : H⁺ , Cl^-

Câu 5 : 

\(n_{NaOH}=n_{HCl}=0.02\cdot0.1=0.002\left(mol\right)\)

\(\Rightarrow x=\dfrac{0.002}{0.01}=0.2\left(M\right)\)

Câu 1 : 

Bảo toàn điện tích : 

\(n_{SO_4^{2-}}=\dfrac{0.2\cdot2+0.1-0.05}{2}=0.225\left(mol\right)\)

\(m_{Muối}=0.2\cdot64+0.1\cdot39+0.05\cdot35.5+0.225\cdot96=40.075\left(g\right)\)

Câu 2 : 

\(\left[Na^+\right]=\dfrac{0.15\cdot0.5\cdot2+0.05\cdot1}{0.15+0.05}=1\left(M\right)\)

1D

2C

3A

5A 6a 7B 8C 9A

10C 11D 12B 13D 14A 15B 16A

17C 18A

19D

20C 21D 22B 23C 24D 25C 26A

27C 28B 29D 30C (V=1lit)

\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)

\(Na+HOOC-COOH\rightarrow NaOOC-COONa+H_2\)

\(0.2................0.2......................................................0.2\)

\(m_{HOOC-COOH}=0.2\cdot90=18\left(g\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

$C_2H_4 + Br_2 \to C_2H_4Br_2$
$n_{Br_2} = n_{C_2H_4} = \dfrac{6,72}{22,4} = 0,3(mol)$
$V_{dd\ Br_2} = \dfrac{0,3}{1} = 0,3(lít) = 300(ml)$

Nếu A không tạo kết tủa thì \(n_{C_2H_2}=n_{C_2Ag_2}=0,2\left(mol\right)\)

=>\(n_A=\dfrac{3}{2}n_{C_2H_2}=0,3\left(mol\right)\)

=> \(m_A=21,4-0,2.26=16,2\)

=> \(M_A=54\)

=> A là C4H6

Vì A không tác dụng với AgNO3/NH3

=> A là \(CH_3-C\equiv C-CH_3\)

=> Chọn C

TH2: A tạo kết tủa 

=> \(n_{C_2H_2}=2x;n_A=3x\)

=> \(m_X=26.2x+3x\left(14n-2\right)=21,4\)

\(m\downarrow=240.2x+3x.\left(14n+105\right)=48\)

=> x=0,036; n.x=0,047

=> n=13,25 (loại)

Ta có : \(n_{HNO_3}=4n_{NO}=0,18\left(mol\right)\)

=> \(n_{NO}=0,045\left(mol\right)\Rightarrow V_{NO}=1,008\left(l\right)\)

\(n_{NO_3^-\left(taomuoi\right)}=3n_{NO}=0,135\left(mol\right)\)

\(m_{muối}=m_{Zn}+m_{NO_3^-\left(taomuoi\right)}+m_{SO_4^{2-}}=32,52\left(g\right)\)

$n_{H_3PO_4} = 0,05.0,5 = 0,025(mol)$
$3NaOH + H_3PO_4 \to Na_3PO_4 + 3H_2O$

Theo PTHH : 

$n_{NaOH} = 3n_{H_3PO_4} = 0,075(mol)$
$V_{dd\ NaOH} = \dfrac{0,075}{1} = 0,075(lít) = 75(ml)$

a) $ K_2HPO_4 + H_3PO_4 \to 2KH_2PO_4$

b) $H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$
c) $H_3PO_4 + Ca(OH)_2 \to CaHPO_4 + 2H_2O$

d) $2H_3PO_4 + 3Ca(OH)_2 \to Ca_3(PO_4)_2 + 6H_2O$