a) Theo đề bài ta có: \(n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\)

PTHH:

\(2Fe+6HCl\rightarrow2FeCl_3+3H_2\uparrow\)

0,1mol....0,3mol.....0,1mol.......0,15mol

a) Ta có: \(m_{Fe}=0,1.56=5,6\left(g\right)\)

b) Ta có: \(m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\)

c) PTHH:

\(4H_2+Fe_3O_4\underrightarrow{t^0}3Fe+4H_2O\)

0,15mol..............0,1125mol

\(\Rightarrow m_{Fe\left(thuduoc\right)}=0,1125.56=6,3\left(g\right)\)

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

\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)

ti le        1      :    2       :      1      :    1

n(mol)     0,5-->1--------->0,5------>0,5

\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)

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

b, Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, Theo PT: \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

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

\(1mol\)     \(2mol\)                 \(1mol\)

\(0,2mol\)    \(0,4mol\)          \(0,2mol\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(m_{HCl}=n.M=0,4.=14,6\left(g\right)\)

a) \(PTHH:Fe+HCL\) → \(FeCl_2+H_2\)

Cân bằng: \(Fe+2HCl\) → \(FeCl_2+H_2\)

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

\(n_{HCl}=2.n_{Fe}=2.0,1=0,2\left(mol\right)\)

\(m_{HCl}=n.M=0,2.36,5=7,3\left(g\right)\)

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

\(V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\)

a) Zn + 2HCl --> ZnCl₂ + H₂

Phản ứng thế

b) \(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

_____0,02------------>0,02-->0,02

=> m_ZnCl2 = 0,02.136 = 2,72(g)

=> V_H2 = 0,02.22,4 = 0,448(l)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\)

Phản ứng thế

\(b,n_{Zn}=\dfrac{1,3}{65}=0,02(mol)\\ \Rightarrow n_{ZnCl_2}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{ZnCl_2}=0,02.136=2,72(g)\\ V_{H_2}=0,02.22,4=0,448(l)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)

Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít

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

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

          0,1-->0,2------------------>0,1

=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)

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3) Cho 6 gam Mg phản ứng 2,24 lít khí oxi(đktc).Sau phản ứng thu được magie oxit(MgO)

a) viết phường trình hóa học

2Mg + O₂ → 2MgO

b) tính khối lượng MgO được tạo thành

m_O2 = 2,24/ 22,4 . 16 = 1,6(g)

m_MgO = m_O2 + m_Mg = 1,6 + 6 = 7,6(g)