Fe+2HCl->FeCl₂+H₂

0,1---------------------0,1

2H2+O₂-to>2H₂O

0,1----0,05 mol

0,1--0,1

n _Fe=\(\dfrac{5,6}{56}\)=0,1 mol

=>V_H2=0,1.22,4=2,24l

=>m_kk=0,05.29=1,45l

Không đủ dữ kiện tính $m_{kk}$ mà chỉ tính được $V_{kk}$

$a)PTHH:Fe+2HCl\to FeCl_2+H_2\uparrow$

$\to$ Phản ứng trao đổi

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

Theo PT: $n_{H_2}=n_{Fe}=0,1(mol)$

$\Rightarrow V_{H_2(đktc)}=0,1.22,4=2,24(lít)$

$c)PTHH:2H_2+O_2\xrightarrow{t^o}2H_2O$

Theo PT: $n_{O_2}=\dfrac{1}{2}n_{H_2}=0,05(mol)$

$\Rightarrow V_{O_2(đktc)}=0,05.22,4=1,12(lít)$

Mà $V_{O_2}$ chiếm $20\%$ của $V_{kk}$

$\Rightarrow V_{kk(đktc)}=\dfrac{1,12}{20\%}=5,6(lít)$

a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

            0,1      0,2                  0,1

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

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

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

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

0,1      0,2                       0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{HCl}=0,2\cdot36,5=7,3g\)

\(2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{CuO}=n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right);n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\\ V_{kk\left(đktc\right)}=\dfrac{100.1,12}{20}=5,6\left(l\right)\\ b,m_{CuO}=0,1.80=8\left(g\right)\\ c,2R+O_2\rightarrow\left(t^o\right)2RO\\ n_R=2.n_{O_2}=2.0,05=0,1\left(mol\right)\\ M_R=\dfrac{2,4}{0,1}=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow R:Magie\left(Mg=24\right)\)

Gọi: \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_X=b\left(mol\right)\end{matrix}\right.\) ⇒ 56a + M_X.b = 21,4 (1)

- Đốt A trong oxi.

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(2X+O_2\underrightarrow{t^o}2XO\)

Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}+\dfrac{1}{2}n_X=\dfrac{2}{3}a+\dfrac{1}{2}b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)

- Cho pư với HCl

+ TH1: X không pư với HCl.

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)=a\)

Thay vào (2) được b = -1/15 → vô lý.

+ TH2: X có pư với HCl.

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(X+2HCl\rightarrow XCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{Fe}+n_X=a+b=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(3\right)\)

Từ (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}a=0,15\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\). Thay vào (1) được M_X = 65 (g/mol)

Vậy: X là kẽm. (Zn)

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5mol\)

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

0,25    0,5                      0,5  ( mol )

\(m_{Fe}=0,25.56=14g\)

\(V_{H_2}=0,5.22,4=11,2l\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,5                          0,5  ( mol )

\(m_{H_2O}=0,5.18=9g\)

\(n_P=\dfrac{m}{M}=\dfrac{12,4}{31}=0,4\left(mol\right)\\ PTHH:4P+5O_2-^{t^o}>2P_2O_5\)

Tỉ lệ          4  :      5      :      2

n(mol)      0,4--->0,5----->0,2

\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\\ V_{kk}=11,2:\dfrac{1}{5}=56\left(l\right)\)

\(m_{P_2O_5}=n\cdot M=0,2\cdot142=28,4\left(g\right)\)

\(PTHH:P_2O_5+3H_2O->2H_3PO_4\)

tỉ lệ            1     :     3       :          2

n(mol)        0,2----->0,6--------->0,4

\(m_{H_3PO_4}=n\cdot M=0,4\cdot98=39,2\left(g\right)\)

\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{39,2}{200}\cdot100\%=19,6\%\)

\(n_K=\dfrac{39}{39}=1\left(mol\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{KOH}=n_K=1\left(mol\right)\\ C_{MddKOH}=\dfrac{1}{0,2}=5\left(M\right)\\ c,2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

a/ PTHH

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b/

Ta có: \(n_{Al_2O_3}=\dfrac{10.2}{102}=0.1\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

4                   2

x                    0.1

\(=>x=\dfrac{0.1\cdot4}{2}=0.2=n_{Al}\)  

\(=>m_{Al}=0.2\cdot27=5.4\left(g\right)\)

Ta có: \(n_{Mg}=\dfrac{14,4}{24}=0,6\left(mol\right)\)

a, PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

_____0,6____________________0,6 (mol)

\(\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)

b, Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}=\dfrac{0,6}{3}\), ta được pư hết.

Theo PT: \(n_{Fe}=2n_{Fe_2O_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\)

Bạn tham khảo nhé!