Câu 2 : 

$Mg + 2HCl \to MgCl_2 + H_2$
$FeO +2 HCl \to FeCl_2 + H_2O$
$Al(OH)_3 + 3HCl \to AlCl_3 + 3H_2O$

Câu 3 : 

a) $ZnO + 2HCl \to ZnCl_2 + H_2O$

b) $n_{ZnCl_2} = n_{ZnO} = \dfrac{16,2}{81} = 0,2(mol)$
$m_{ZnCl_2} = 0,2.136 = 27,2(gam)$

nNa2O= 12,4/62=0,2(mol)

a) PTHH: Na2O + H2O -> 2 NaOH

0,2_________0,2____0,4(mol)

b) VddNaOH=2(l)

=>CMddNaOH=0,4/0,2=2(M)

Chúc em học tốt!

bạn xem lại xem 13.5(g) hay 13.8g nhé ^^ ,cho tròn số ý mà

CuCl2+2NaOH->Cu(OH)2+2NaCl

nCuCl2=13.5:138=0.1(mol)

nNaOH=20:40=0.5(mol)

theo pthh:nNaOH=2nCuCl2

theo bài ra,nNaOH=5 nCuCl2->NaOH dư tính theo CuCl2

theo pthh,nCu(OH)2=nCuCl2->nCu(OH)2=0.1(mol)

mCu(OH)2=0.1*98=9.8(g)

b)PTHH:Cu(OH)2+2HCl->CuCl2+2H2O

theo pthh:nHCl=2nCu(OH)2->nHCl=0.1*2=0.2(mol)

mHCl=0.2*36.5=7.3(g)

mDD HCl=7.3*100:10=73(g) 

\(a,PTHH:CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\\ Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\)

Hiện tượng: Dung dịch màu xanh nhạt dần, xuất hiện kết tủa màu xanh lơ

Phản ứng phân hủy Cu(OH)₂ sinh ra chất rắn CuO màu đen và nước

\(b,n_{CuCl_2}=\dfrac{13,5}{135}=0,1\left(mol\right)\\ m_{NaOH}=\dfrac{200\cdot2,5\%}{100\%}=5\left(g\right)\\ \Rightarrow n_{NaOH}=\dfrac{5}{40}=0,125\left(mol\right)\)

Vì \(\dfrac{n_{CuCl_2}}{1}>\dfrac{n_{NaOH}}{2}\) nên CuCl₂ dư

\(\Rightarrow\dfrac{1}{2}n_{NaOH}=n_{Cu\left(OH\right)_2}=n_{CuO}=0,0625\left(mol\right)\\ \Rightarrow m=m_{CuO}=0,0625\cdot80=5\left(g\right)\)

\(c,n_{NaCl}=n_{NaOH}=0,125\left(mol\right)\\ \Rightarrow m_{NaCl}=0,125\cdot58,5=7,3125\left(g\right)\\ m_{Cu\left(OH\right)_2}=0,0625\cdot98=6,125\left(g\right)\\ \Rightarrow m_{dd_{Cu\left(OH\right)_2}}=13,5+200-7,3125=206,1875\left(g\right)\\ \Rightarrow C\%_{Cu\left(OH\right)_2}=\dfrac{6,125}{206,1875}\cdot100\%\approx2,97\%\)

\(n_{CO_2}=0,2\left(mol\right)\)

a)\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\) (1)

b)Từ (1)\(\Rightarrow n_{CaCO_3}=n_{CO_2}=0,2mol\)

\(\Rightarrow m_{CaCO_3}=100.0,2=20\left(g\right)\)

c)Từ (1)\(\Rightarrow n_{Ca\left(OH\right)_2}=n_{CO_2}=0,2mol\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(0,2\)                         \(0,2\)        (mol)

\(\Rightarrow n_{CaO}=0,2mol\)

\(\Rightarrow m_{CaO}=11,2\left(g\right)\)

a, \(CaO+Ca\left(OH\right)_2->CaCO_3+H_2O\)

b,\(n_{CaCO_3}=n_{CO_2}=0,2mol\) \(->m_{CaCO_3}=0,2.100=20g\)

c,\(CaO+H_2O->Ca\left(OH\right)_2\)

\(n_{CaO}=n_{Ca\left(OH\right)_2}=0,2mol\) \(->m_{CaO}=56.0,2=11,2g\)

Có sai đề không

Bài1:

a,Vì dd A là dd bazo nên làm cho quỳ tím đổi thành màu xanh

b,\(n_{Na_2O}=\dfrac{21,7}{62}=0,35\left(mol\right)\)

PTHH: Na₂O + H₂O → 2NaOH

Mol:       0,35                     0,7

\(\Rightarrow C_{M_{ddNaOH}}=\dfrac{0,7}{0,4}=1,75M\)

Bài 2:

a,\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:    0,15     0,3                      0,15

⇒ a=m_Zn = 0,15.65 = 9,75 (g)

b,\(V_{HCl}=\dfrac{0,3}{1,5}=0,2\left(l\right)=200\left(ml\right)\)

1

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

a

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,3-->0,6---->0,3------->0,3

b

\(C\%_{dd.HCl}=\dfrac{0,6.36,5.100\%}{400}=5,475\%\)

c

\(m_{dd}=16,8+400-0,3.2=416,2\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{0,3.127.100\%}{416,2}=9,15\%\)

2

\(n_{HCl}=\dfrac{200.7,3\%}{100\%}:36,5=0,4\left(mol\right)\)

a

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,2<--0,4------>0,2------>0,2

b

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c

\(x=m_{Mg}=0,2.24=4,8\left(g\right)\)

d

\(m_{dd}=4,8+200-0,2.2=204,4\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0,2.95.100\%}{204,4}=9,3\%\)

3

\(n_{H_2SO_4}=\dfrac{300.9,8\%}{100\%}:98=0,3\left(mol\right)\)

a

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

0,2<--0,3--------->0,1------------>0,3

b

\(m_{Al}=0,2.27=5,4\left(g\right)\)

c

\(m_{dd}=5,4+300-0,3.2=304,8\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342.100\%}{304,8}=11,22\%\)

Câu 1:

\(C+O_2\xrightarrow{t^o}CO_2\\ CO_2+H_2O\buildrel{{}}\over\rightleftharpoons H_2CO_3\\ H_2CO_3+2NaOH\to Na_2CO_3+2H_2O\\ Na_2CO_3+H_2SO_4\to Na_2SO_4+H_2O+CO_2\uparrow\\ Na_2SO_4+BaCl_2\to 2NaCl+BaSO_4\downarrow\)

a) mH2SO4= 20%.73,5=14,7(g) -> nH2SO4=0,15(mol)

PTHH: CuO + H2SO4 -> CuSO4 + H2O

x______________x____x(mol)

ZnO + H2SO4 -> ZnSO4 + H2O

y_____y_______y(mol)

Ta có: \(\left\{{}\begin{matrix}80x+81y=12,1\\x+y=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

b) mCuO=0,05.80=4(g) 

-> %mCuO= (4/12,1).100=33,058%

=>%mZnO= 66,942%

Bài 3:

n_KOH = 1.0,05 = 0,05 (mol);n_HCl=0,5.0,5=0,25 (mol)

PTHH:   Fe      +      2HCl    →    FeCl₂    +    H₂

Mol:    0,0125                                 0,0125

PTHH: 2KOH + FeCl₂ → 2KCl + Fe(OH)₂

Mol:      0,05      0,025

Ta có:\(\dfrac{0,25}{2}>\dfrac{0,0125}{1}\) ⇒ HCl dư, FeCl₃ pứ hết

⇒ m=0,0125.56 = 0,7 (g)

Bài 4:

a,Nếu cho giấy quỳ tím vào dung dịch sau phản ứng thì màu của giấy quỳ tím sẽ không thay đổi màu sắc 

b,\(n_{NaOH}=0,5.0,15=0,075\left(mol\right);n_{HCl}=1.0,15=0,15\left(mol\right)\)

PTHH: NaOH + HCl → NaCl + H₂O

Mol:      0,075    0,075    0,075

Ta có:\(\dfrac{0,075}{1}< \dfrac{0,15}{1}\) ⇒ NaOH pứ hết , HCl dư

m_NaCl = 0,075.58,5 = 4,3875 (g)

c, V_{dd sau pứ} = 0,15 + 0,15 = 0,3 (l)

\(C_{M_{ddNaCl}}=\dfrac{0,075}{0,3}=0,25M;C_{M_{ddHCldư}}=\dfrac{0,15-0,075}{0,3}=0,25M\)

Bài 5:

a,\(n_{BaCl_2}=\dfrac{100.20,8\%}{208}=0,1\left(mol\right);n_{Na_2SO_4}=\dfrac{150.14,2\%}{142}=0,15\left(mol\right)\)

PTHH: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

Mol:      0,1             0,1                0,1

Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\)⇒ BaCl₂ pứ hết, Na₂SO₄ dư

\(\Rightarrow m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

b,m_{dd sau pứ} = 100+150 = 250 (g)

\(C\%_{ddNaCl}=\dfrac{0,2.58,5.100\%}{250}=4,68\%\)

 \(C\%_{ddNa_2SO_4dư}=\dfrac{\left(0,15-0,1\right).142.100\%}{250}=2,84\%\)

Bài 2:

a,\(n_{CaCO_3}=\dfrac{35}{100}=0,35\left(mol\right)\)

PTHH: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Mol:       0,35           0,7                            0,35

\(\Rightarrow V_{CO_2}=0,35.22,4=7,84\left(l\right)\)

b,\(C_{M_{ddHCl}}=\dfrac{0,7}{0,5}=1,4M\)