Fe+2HCl->FeCl2+H2

0,125---0,25--0,125----0,125---

n Fe=11.2\56=0,2 mol

n HCl=0,25.1=0,25 mol

=> lập tỉ lệ : 0,2\1>0,25\2

=>HCl hết

=>VH2=0,125.22,4=2,8l

=>m Fe=0,125.56=7g

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

\(n_{HCl}=CM.V_{dd}=1.0,25=0,25\left(mol\right)\)

PTHH:\(2Fe+6HCl\rightarrow2FeCl_3+3H_2\)

TPƯ:   0,2       0,25

PƯ:     0,08    0,25         0,08         0,125

SPƯ:   0,12        0            0,08       0,125

\(V_{H_2}=n.22,4=0,125.22,4=2,8\left(l\right)\)

\(m_{Fedư}=n.M=0,12.56=6,72\left(g\right)\)

a, \(n_{Cu}=\dfrac{9,6}{64}=0,15\left(mol\right)\)

PTHH: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

Mol:     0,15                          0,15            0,1

\(V_{NO}=0,1.22,4=2,24\left(l\right)\)

b, \(C_{M_{ddCu\left(NO_3\right)_2}}=\dfrac{0,15}{0,2}=0,75M\)

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

Ta có: \(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,3\left(mol\right)\\n_{ZnCl_2}=0,15\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{ZnCl_2}=0,15\cdot136=20,4\left(g\right)\\C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)

Ta có: \(n_{Cu\left(NO_3\right)_2}=0,2.1,5=0,3\left(mol\right)\)

PT: \(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_{2\downarrow}+2NaNO_3\)

_______0,3_______0,6_______0,3_________0,6 (mol)

a, m_Cu(OH)2 = 0,3.98 = 29,4 (g)

b, \(V_{ddNaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

c, \(C_{M_{NaNO_3}}=\dfrac{0,6}{0,2+0,3}=1,2M\)

Bạn tham khảo nhé!

a) \(n_{Cu\left(NO_3\right)_2}=1,5.0,2=0,3\left(mol\right)\)

\(Cu\left(NO_3\right)_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)

\(n_{Cu\left(OH\right)_2}=n_{Cu\left(NO_3\right)_2}=0,3\left(mol\right)\)

=> \(m_{Cu\left(OH\right)_2}=29,4\left(g\right)\)

b) \(n_{NaOH}=2n_{Cu\left(OH\right)_2}=0,6\left(mol\right)\)

=> \(V_{NaOH}=\dfrac{0,6}{2}=0,3\left(l\right)\)

c) \(CM_{NaCl}=\dfrac{0,3.2}{0,2+0,3}=1,2M\)

a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,2                              0,2

(do Cu ko tác dụng với HCl loãng)

b, \(m_{Zn}=0,2.65=13\left(g\right)\)

  \(m_{Cu}=19,4-13=6,4\left(g\right)\)

PTHH: \(X_aO_b+bH_2SO_4\rightarrow X_a\left(SO_4\right)_b+bH_2O\)

Giả sử \(n_{H_2SO_4}=1\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=490\left(g\right)\)

Theo PTHH: \(n_{X_aO_b}=n_{X_a\left(SO_4\right)_b}=\dfrac{1}{b}\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{X_a\left(SO_4\right)_b}=\dfrac{a.X}{b}+96\left(g\right)\\m_{X_aO_b}=\dfrac{a.X}{b}+16\left(g\right)\end{matrix}\right.\)

Theo đề: \(\dfrac{\dfrac{aX}{b}+96}{\dfrac{aX}{b}+506}=0,2264\) \(\Rightarrow\dfrac{0,7736aX}{b}=18,5584\) \(\Rightarrow\dfrac{aX}{b}\approx24\)

  Với \(a=b\ne0\) thì \(X=24\) (Magie)

  Vậy công thức oxit là MgO

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

b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)

⇒ m_Zn = 0,2.65 = 13 (g)

⇒ m_Cu = 19,4 - 13 = 6,4 (g)

Bạn tham khảo nhé!

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

\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)

        1                                                  0,5       ( mol )

\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\)

       1                                                    1                       ( mol )

\(m_{CH_3COOH}=1.60.80\%=48g\)