a) n Fe = 16,8/56 = 0,3(mol)

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$

n H2 = 3/2 n Fe = 0,45(mol)

=> V H2 = 0,45.22,4 = 10,08(lít)

b)

$Fe + 2HCl \to FeCl_2 + H_2$

n HCl = 2n Fe = 0,6(mol)

=> m HCl = 0,6.36,5 = 21,9 gam

a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,1-->0,2------>0,1-->0,1

=> V_H2 = 0,1.22,4 = 2,24 (l)

m_ZnCl2 = 0,1.136 = 13,6 (g)

b) \(C\%_{dd.HCl}=\dfrac{0,2.36,5}{200}.100\%=3,65\%\)

c) \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,15}{1}\) => H₂ hết, O₂ dư

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,1--------------->0,1

=> m_H2O = 0,1.18 = 1,8 (g)

a) \(n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\)

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

            \(\dfrac{3}{14}\)---------------------->\(\dfrac{3}{14}\)

\(\Rightarrow V_{H_2}=\dfrac{3}{14}.22,4=4,8\left(l\right)\)

b) \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

PTHH: \(ZnO+H_2\xrightarrow[]{t^o}Zn+H_2O\)

Xét tỉ lệ: \(0,1< \dfrac{3}{14}\Rightarrow H_2\) dư

Theo PT: \(n_{Zn}=n_{ZnO}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

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

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

\(\dfrac{3}{14}mol\)                        \(\dfrac{3}{14}mol\)

\(a)n_{Fe}=\dfrac{m}{M}=\dfrac{12}{56}\approx0,21=\dfrac{3}{14}\left(mol\right)\)

\(V_{H_2}=n.22,4=\dfrac{3}{14}.22,4=4,8\left(l\right)\)

\(b)n_{ZnO}=\dfrac{m}{M}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

\(ZnO+H_2\rightarrow Zn+H_2O\)

\(1mol\)    \(1mol\)    \(1mol\)

\(0,1mol\)  \(0,1mol\)  \(0,1mol\)

\(\text{Ta thấy }H_2\text{ dư,ZnO phản ứng hết.Bài toán tính theo ZnO}\)

\(m_{Zn}=n.M=0,1.65=6,5\left(g\right)\)

2Al + 2H₂O + 2NaOH→ 3H₂ + 2NaAlO₂

0,2mol 0,3mol

mAl=0,2.27=5,4g

2Al + 6HCl→ 2AlCl₃+ 3H₂

0,2mol 0,3mol

Fe + 2HCl→ FeCl₂+ H₂

0,15mol 0,45-0,3 mol

mFe=0,15.56=8,4g

mCu=32,8-(6,4+8,4)=18g

%mFe=\(\frac{8,4}{32,8}.100=25,6\%\)

%mCu=\(\frac{18}{32,8}.100=54,8\%\)

%mAl=19,6%

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

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: 56n_Fe + 65n_Zn = 35,4 (1)

Theo PT: \(n_{H_2}=n_{Fe}+n_{Zn}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,4\left(mol\right)\\n_{Zn}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,4.56=22,4\left(g\right)\\m_{Zn}=0,2.65=13\left(g\right)\end{matrix}\right.\)

a,\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:    0,15     0,3

Ta có: \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\) ⇒ H₂ pứ hết,Fe dư

\(V_{H_2}=3,36\left(l\right)\) (đề cho)

b, ko tính đc k/lg dd ,chỉ tính đc thể tích dd

\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\) 

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\) 

                1        2               1           1

             0,05      0,1          0,05       0,05

a) \(V_{H_2}=n.24,79=0,05.24,79=1,2395\left(l\right)\) 

\(m_{ZnCl_2}=n.M=0,05.\left(65+35,5.2\right)=6,8\left(g\right)\) 

b) \(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 

Ta cos tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\Rightarrow\) CuO dư.

Theo ptr, ta có: \(n_{Cu}=n_{H_2}=0,05mol\\ \Rightarrow m_{Cu}=n.M=0,05.64=3,2\left(g\right).\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

Ta có: \(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

THeo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,05\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,05.24,79=1,2395\left(l\right)\)

\(m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)

b, Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{H_2}=0,05\left(mol\right)\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)\)

\(a.n_{Zn}=\dfrac{13}{65}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(V_{H_2}=0,2.22,4=4,48l\\ b.m_{ddHCl}=\dfrac{0,4.36,5}{20}\cdot100=73g\\ c.CuO+H_2\underrightarrow{t^0}Cu+H_2O\\ n_{Cu}=n_{H_2}=0,2mol\\ m_{Cu}=0,2.64=12,8g\)