Fe+2HCl->FeCl₂+H₂

0,1----0,2-------------0,1 mol

n _Fe=5,6\56=0,1 mol

=>m _HCl=0,2.36,5=7,3g

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

\(a,PTHH:KHCO_3+2H_2SO_4\rightarrow K_2SO_4+2CO_2\uparrow+2H_2O\\ b,n_{KHCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ \Rightarrow n_{CO_2}=2n_{KHCO_3}=0,4\left(mol\right)\\ \Rightarrow V_{CO_2}=0,4\cdot22,4=8,9\left(l\right)\\ c,n_{H_2SO_4}=n_{CO_2}=0,4\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,4\cdot98=39,2\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{39,2\cdot100\%}{19,6\%}=200\left(g\right)\)

Bài 1:

A . PTHH : 2AI + 3H₂SO₄ => AI₂(SO₄)₃ + 3H₂

B . nAI = 7.1 : 27 = 0.2 (mol)

 PT :2AI         +       3H₂SO₄      =>        AI₂(SO₄)₃       +   3H₂

        _{2 mol              3 mol                                                       3 mol}

        _{0.2 mol            0.3 mol                                                    0.3 mol}

=> V_{H2 = 22.4 X 0.3 = 6.72 (lít)}  

3.  500ml = 0.5 lít

Nồng độ mol/l của dd H₂SO₄ là:

C_{M H2SO4} = mol/lít =0.3/0.5 = 0.6 MOL/ LÍT

chuẩn nha hoàng kiều anh

a) 

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

b)  \(n_{Zn=\frac{16,12}{65}=0,248\left(mol\right)}\)

Theo PTHH :  \(n_{H_2}=n_{Zn}=0,248\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=0,248\times22,4=5,5552\left(l\right)\)

c ) Ta có :

Theo PTHH :  \(n_{HCl}=2n_{Zn}=0,496\left(mol\right)\)

\(\Rightarrow m_{HCl}=18,104\left(g\right)\)

\(m_{dd}=\frac{18,104\times100}{10}=181,04\left(g\right)\)

a) \(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\) 

b) \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\) 

\(n_{HCl}=2.n_{Mg}=0,2.2=0,4mol\)

\(\Rightarrow m_{HCl}=n.M=0,4.36,5=14,6g\)

c) \(n_{H_2}=n_{Mg}=0,2mol\) 

Thể tích khí hidro sinh ra (ở đktc): 

\(V_{H_2}=0,2.24,79=4,958l.\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

b, Ta có: 24n_Mg + 56n_Fe = 10,4 (1)

Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(2\right)\)

Từ (1)  và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2.24=4,8\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

a) Mg + 2HCl --> MgCl₂ + H₂

b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             0,3<--0,6<-----------0,3

=> m_Mg = 0,3.24 = 7,2 (g)

=> m_Ag = 10,4 - 7,2 = 3,2 (g)

c) \(V_{dd.HCl}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

0,3                                     0,3

\(m_{Mg}=0,3\cdot24=7,2g\Rightarrow m_{Ag}=10,4-7,2=3,2g\)

\(n_{HCl}=0,6mol\Rightarrow V_{HCl}=\dfrac{0,6}{0,5}=1,2l\)

a) PTHH: Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,2-->0,4------>0,2-->0,2

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

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

d) \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ n_{H_2}=n_{Fe}=0,25\left(mol\right)\\ V_{H_2}=0,25.22,4=5,6\left(l\right)\\ n_{HCl}=2n_{Fe}=0,5\left(mol\right)\\ m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(n_{Fe}=\dfrac{22,4}{56}=0,4\) (mol) (1)

Phương trình hóa học : 

Fe + 2HCl ---> FeCl₂ + H₂ (2) 

Từ (1) và (2) ta có \(n_{FeCl_2}=n_{H_2}=0,4\) (mol) ; \(n_{HCl}=0,8\left(mol\right)\)

b) => \(m_{\text{muối}}=0,4.\left(56+35,5.2\right)=50.8\left(g\right)\)

c) \(V_{\text{khí}}=0,4.22,4=8,96\left(l\right)\)

d) \(m_{HCl}=0,8.36.5=29,2\left(g\right)\)

\(\Rightarrow C\%=\dfrac{29,2}{200}.100\%=14,6\%\)