Mg+    2HCl→       MgCl₂+    H₂

(mol)   0,1          0,2            0,1          0,1                                             

a) \(n_{Mg}=\dfrac{m}{M}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

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

b) Đổi: 100ml=0,1 lít

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

c) \(m_{MgCl_2}=n.M=0,1.95=9,5\left(g\right)\)

Câu `3:`

`n_[Mg]=[2,4]/24=0,1(mol)`

`Mg + 2HCl -> MgCl_2 + H_2 \uparrow`

`0,1`      `0,2`               `0,1`      `0,1`       `(mol)`

`a)C%_[HCl]=[0,2.36,5]/200 . 100=3,65%`

`b)m_[MgCl_2]=0,1.95=9,5(g)`

`c)V_[H_2]=0,1.22,4=2,24(l)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Câu `4:`

`n_[Zn]=[3,25]/65=0,05(mol)`

`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`

`0,05`  `0,1`            `0,05`    `0,05`       `(mol)`

`a)C%_[HCl]=[0,1.36,5]/200 .100=1,825%`

`b)m_[ZnCl_2]=0,05.136=6,8(g)`

`c)V_[H_2]=0,05.22,4=1,12(l)`

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=0,2.1,5=0,3\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Mg hết, H₂SO₄ dư

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

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

=> V_H2 = 0,2.22,4 = 4,48 (l)

\(\left\{{}\begin{matrix}C_{M\left(MgSO_4\right)}=\dfrac{0,2}{0,2}=1M\\C_{M\left(H_2SO_4.dư\right)}=\dfrac{0,3-0,2}{0,2}=0,5M\end{matrix}\right.\)

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

a, \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

b, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

c, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{MgCl_2}=n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,4.24=9,6\left(g\right)\)

c, \(m_{MgCl_2}=0,4.95=38\left(g\right)\)

d, Bạn bổ sung thêm thể tích dd HCl nhé.

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

PTHH: 2A + 2H₂O --> 2AOH + H₂

           0,1<------------0,1<--0,05

=> \(M_A=\dfrac{2,3}{0,1}=23\left(g/mol\right)\)

=> A là Na (Natri)

\(C_{M\left(ddNaOH\right)}=\dfrac{0,1}{0,2}=0,5M\)

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

PTHH: 2A + 2H₂O --> 2AOH + H₂

           0,2<-------------0,2<----0,1

=> \(M_A=\dfrac{7,8}{0,2}=39\left(g/mol\right)\)

=> A là Kali (K)

n_KOH = 0,2 (mol)

=> \(C_{M\left(X\right)}=\dfrac{0,2}{0,2}=1M\)

a/ PTHH: R + 2HCl ===> RCl₂ + H₂

n_H2 = 2,688 / 22,4 = 0,12 (mol)

n_R = nH2 = 0,12 mol

=> M_R = 6,72 / 0,12 = 56 (g/mol)

=> R là Fe