a)

Fe+ 2HCl→ FeCl2+H2↑

b)

nH2= 0,15 mol

⇒ nFe= 0,15 mol⇒ mFe= 0,15.56=8,4 g

c)\(CM_{HCl}=\frac{0,15.2}{0,15}\text{ = 2M}\)

Fe+ H₂SO₄\(\rightarrow\)FeSO₄+H₂

\(n_{Fe}=n_{H_2SO_4}=n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

m_Fe=0,1.56=5,6g

\(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5M\)

a.Fe+H2SO4->FeSO4+H2

nH2=2,24/22,4=0,1(mol)

=>nFe=0,1(mol)=>mFe=0,1*56=5,6(g)

b.nH2SO4=0,1(mol)

=>CM_H2SO4=0,1/0,2=0,05(M)

PTHH: Fe+ H2SO4-> FeSO4 + H2

a) nFe = n_H2SO4 = n_H2

--> n_H2 = \(\dfrac{2,24}{22,4}\) = 0,1 mol

--> m_Fe = 0,1. 56 = 5,6 (g)

b) C_M(H2SO4) = \(\dfrac{n}{V}\) = \(\dfrac{0,1}{0,2}\) = 0,5M

Fe + H₂SO₄ → FeSO₄ + H₂ (1)

\(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=0,5\times0,2=0,1\left(mol\right)\)

Theo PT1: \(n_{Fe}=n_{H_2SO_4}\)

Theo bài: \(n_{Fe}=n_{H_2SO_4}\)

Vì \(1=1\) ⇒ phản ứng vừa đủ

a) Theo PT1: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)

b) Theo PT1: \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{FeSO_4}}=\frac{0,1}{0,5}=0,2\left(M\right)\)

c) Ta có: \(n_{FeSO_4.7H_2O}=n_{FeSO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeSO_4.7H_2O}=0,1\times278=27,8\left(g\right)\)

d) 4H₂ + Fe₃O₄ \(\underrightarrow{to}\) 3Fe + 4H₂O (2)

Theo Pt2: \(n_{Fe_3O_4}=\frac{1}{4}n_{H_2}=\frac{1}{4}\times0,1=0,025\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,025\times232=5,8\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

_____0,05__0,1____________0,05 (mol)

b, m_Fe = 0,05.56 = 2,8 (g)

c, m_HCl = 0,1.36,5 = 3,65 (g)

\(\Rightarrow m_{ddHCl}=\dfrac{3,65}{10\%}=36,5\left(g\right)\)

Bạn tham khảo nhé!

\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

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

\(0.05......0.1...................0.05\)

\(m_{Fe}=0.05\cdot56=2.8\left(g\right)\)

\(m_{dd_{HCl}}=\dfrac{0.1\cdot36.5\cdot100}{10}=36.5\left(g\right)\)

a. PTHH : Fe + 2HCl -> FeCl₂ + H₂ 

b. Ta có PTHH : Fe + 2HCl -> FeCl₂ + H₂ 

Theo pt             1mol  2mol

Theo đề          0,1mol->0,2mol

Khối lượng HCl đã phản ứng :

\(m_{HCl}\) = \(n_{HCl}\) . \(M_{HCl}\)

=> \(m_{HCl}\) = 0,2 . 36,5

=>\(m_{HCl}\) = 7,3 (g)

a, Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

a--->2a------------------>a

2Al + 6HCl ---> 2AlCl₃ + 3H₂

b---->3b-------------------->1,5b

=> \(\left\{{}\begin{matrix}56a+27b=16,6\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow a=b=0,2\left(mol\right)\)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) \(C\%_{HCl}=\dfrac{\left(0,2.2+0,2.3\right).36,5}{300}.100\%=12,167\%\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) 
gọi nFe : a , nAl: b (a,b>0)  => 56a + 27b = 16,6 (g) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
           a                                     a
         \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\) 
           b                                     \(\dfrac{3b}{2}\)
          
=> \(a+\dfrac{3b}{2}=0,5\) 
ta có hệ pt 
\(\left\{{}\begin{matrix}56a+27b=16,6\\a+\dfrac{3b}{2}=0,5\end{matrix}\right.\) 
=> a= 0,2 , b = 0,2 
\(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=16,6-11,2=5,4\left(g\right)\end{matrix}\right.\) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,2     0,4 
        \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,2    0,6 
=> \(m_{HCl}=\left(0,4+0,6\right).36,5=36,5\left(g\right)\) 
=> \(C\%=\dfrac{36,5}{200}.100\%=18,25\%\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,1--------------->0,1---->0,1

=> m_FeCl2  = 0,1.127 = 12,7(g)

c) V_H2 = 0,1.22,4 = 2,24(l)

\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{FeCl_2}=n_{H_2}=0,1(mol)\\ a,m_{FeCl_2}=0,1.127=12,7(g)\\ b,V_{H_2}=0,1.22,4=2,24(l)\)

Câu 1: Đặt CÔng thức Fe_xO_y

Fe_xO_y+yCO\(\rightarrow\)xFe+yCO₂

CO₂ bị Ca(OH)₂ hấp thụ nên m=\(m_{CO_2}\)

Fe+2HCl\(\rightarrow\)FeCl₂+H₂

\(n_{Fe}=n_{H_2}=0,1mol\)

Theo PTHH 1 ta có: noxit\(=\dfrac{1}{x}.n_{Fe}=\dfrac{0,1}{x}mol\)

MOxit\(=\dfrac{m}{n}=\dfrac{8}{\dfrac{0,1}{x}}=80x\)

Hay 56x+16y=80x suy ra 24x=16y hay 3x=2y\(\Leftrightarrow\dfrac{x}{y}=\dfrac{2}{3}\)

Fe₂O₃

\(n_{CO_2}=\dfrac{y}{x}n_{Fe}=\dfrac{3}{2}.0,1=0,15mol\)

m=0,15.44=6,6g

Cu+2AgNO₃\(\rightarrow\)Cu(NO₃)₂+2Ag

-Gọi số mol Cu là x theo PTHH số mol Ag là 2x

- Độ tăng khối lượng=108.2x-64x=7,6

152x=7,6 hay x=0,05mol

\(n_{AgNO_3}=2n_{Cu}=2.0,05=0,1mol\)

\(m_{AgNO_3}=0,1.170=17g\)

\(m_{AgNO_3}=0,1.170=17g\)

\(m_{ddAgNO_3}=v.D=400.1,05=420g\)

\(C\%=\dfrac{17.100}{420}\approx4,05\%\)

mdd=420-7,6=412,4g

\(n_{Cu\left(NO_3\right)_2}=n_{Cu}=0,05mol\)

\(m_{Cu\left(NO_3\right)_2}=0,05.188=9,4g\)

\(C\%=\dfrac{9,4.100}{412,4}\approx2,3\%\)