\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)

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

\(b.\)

\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)

\(m_{Mg}=0.25\cdot24=6\left(g\right)\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(c.\)

\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)

Bài 1

\(a)n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,15      0,3          0,15         0,15

\(\%m_{Fe}=\dfrac{0,15.56}{12}\cdot100\%=70\%\\ \%m_{FeO}=100\%-70\%=30\%\\ b)n_{FeO}=\dfrac{12-0,15.56}{72}=0,05mol\\ FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05        0,1

\(V_{ddHCl}=\dfrac{0,1+0,3}{2}=0,2l\)

Bài 2

\(a)Na_2O+H_2O\rightarrow2NaOH\\ b)BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ c)BaSO_4?\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+CuSO_4\rightarrow BaSO_4+Cu\left(OH\right)_2\\ d)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ e)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\)

a, \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

b,\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Na_2CO_3}=0,4\left(mol\right)\)

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

c, \(n_{CO_2}=n_{Na_2CO_3}=0,2\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

Mg+2HCl->MgCl2+H2

0,3---0,6--------------0,3 mol

Cu ko td với HCl

n H2=6,72\22,4=0,3 mol

=>m Mg=0,3.24=7,2g

=>m Cu=13,6-7,2=6,4g

=>VHCl=0,6\1=0,6l=600ml

a+b) PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{CO_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\\V_{CO_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)

c) Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=\dfrac{50\cdot40\%}{40}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\) Tạo muối trung hòa, bazơ dư, tính theo CO₂

Bảo toàn Cacbon: \(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\) \(\Rightarrow m_{Na_2CO_3}=0,1\cdot106=10,6\left(g\right)\)

a) $Fe +2HCl \to FeCl_2 + H_2$

b) $n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)$
$V_{H_2} = 0,1.24,79 = 2,479(lít)$

c) $n_{HCl} = 2n_{Fe} = 0,2(mol)$
$C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M$

nFe=0,1 mol

Fe              +2HCl=>FeCl2+H2

0,1 mol=>0,2 mol          =>0,1 mol

VH2=0,1.22,4=2,24 lít

nHCl=0,2 mol=>mHCl=0,2.36,5=7,3g

=>C% dd HCl=7,3/200.100%=3,65%

a ,\(Zn+2HCl=>ZnCl_2+H_2\) (1)

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

theo (1) \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\) 

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

c, Theo (1) \(n_{HCl}=2n_{Zn}=0,2\left(mol\right)\)  

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

nồng độ % dung dịch axit đã dùng là

\(\frac{7,3}{200}.100\%=36,5\%\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{8,4}{56}=0,15mol\)

PTHH: Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

 TL:        1        2           1         1

mol:     0,15 \(\rightarrow\) 0,3 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15

Đổi \(100ml=0,1l\)

\(b.C_{M_{ddHCl}}=\dfrac{n}{V_{dd}}=\dfrac{0,3}{0,1}=3M\)

\(c.V_{H_2}=n.22,4=0,15.22,4=33,6l\)

d. Ta có: \(n_{H_2}=0,15mol\)

PTHH: H₂ + CuO \(\rightarrow\) Cu + H₂O

  TL:     1         1          1        1

mol:  0,15 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15

\(n_{CuO}=\dfrac{m}{M}=\dfrac{20}{80}=0,25mol\)

Lập tỉ lệ: \(\dfrac{n_{H_2}}{1}:\dfrac{n_{CuO}}{1}\)

      \(\Leftrightarrow=\dfrac{0,15}{1}< \dfrac{0,25}{1}\)

\(\Rightarrow\) H₂ hết, CuO dư \(\Rightarrow\) Tính theo H₂

\(m_{CuO}=n.M=0,15.64=9,6g\)