Câu 1:

\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ a,PTHH:CO_2+Ba(OH)_2\to BaCO_3\downarrow+H_2O\\ \Rightarrow n_{Ba(OH)_2}=n_{BaCO_3}=n_{CO_2}=0,5(mol)\\ \Rightarrow C_{M_{Ba(OH)_2}}=\dfrac{0,5}{0,2}=2,5M\\ m_{BaCO_3}=0,5.197=98,5(g)\\ b,PTHH:Ba(OH)_2+2HCl\to BaCL_2+2H_2O\\ \Rightarrow n_{HCl}=2n_{Ba(OH)_2}=1(mol)\\ \Rightarrow m_{CT_{HCl}}=1.36,5=36,5(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{36,5}{20\%}=182,5(g)\)

Câu 2:

\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2(mol)\\ m_{HCl}=\dfrac{292.20\%}{100\%}=58,4(g)\\ \Rightarrow n_{HCl}=\dfrac{58,4}{36,5}=1,6(mol)\\ PTHH:Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\)

Vì \(\dfrac{n_{HCl}}{6}>\dfrac{n_{Fe_2O_3}}{1}\) nên \(HCl\) dư

\(\Rightarrow n_{FeCl_3}=2n_{Fe_2O_3}=0,4(mol);n_{H_2O}=3n_{Fe_3O_3}=0,6(mol)\\ \Rightarrow \begin{cases} m_{CT_{FeCl_3}}=0,4.162,5=65(g)\\ m_{H_2O}=0,6.18=10,8(g) \end{cases}\\ \Rightarrow m_{dd_{FeCl_3}}=32+292-10,8=313,2(g)\\ \Rightarrow C\%_{FeCl_3}=\dfrac{65}{313,2}.100\%\approx20,75\%\)

`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`

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

`n_[Fe]=[5,6]/56=0,1(mol)`

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

`b)C_[M_[HCl]]=[0,2]/[0,1]=2(M)`

n_Al = 5.4/27 = 0.2  (mol) 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2____0.3_________________0.3 

VH2 = 0.3*22.4 = 6.72(l) 

CM H2SO4 = 0.3/0.1 = 3 M

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) 

\(\Rightarrow n_{H_2}=0,3mol=n_{H_2SO_4}\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\C_{M_{H_2SO_4}}=\dfrac{0,3}{0,1}=3\left(M\right)\end{matrix}\right.\)

a.

m_AgNO3 = (200.8,5%)/100 =17g

n_AgNO3  = 17/170= 0,1 mol

để kết tủa hoàn toàn thì n_AgNO3=n_HCl = 0,1 mol

C_HCl=0,1/0,15=2/3 (M)

b.

HCl + NaHCO₃   =====> NaCl + CO₂ + H₂O

n_CO2 = 0,1 mol=n_HCl    (theo pt)

m_HCl =3,65 g

%C_HCl = (3,65/50) .100% =7,3%

Có n_Zn= 0,65/65=0,1(mol)

a,PTPƯ: Zn + 2HCl ---> ZnCl₂ + H₂

(mol) 0,1 ---> 0,05 ---> 0,1 ---> 0,1

b, Theo pt có m_muối= 0,1.136=13.6(g)

Lại có V_khí=0,1.22,4= 2,24(l)

c, Theo pt có m_HCl= 0,05.36,5=1,825(g)

=> C%=\(\dfrac{1,825}{200}\).100%=0,9125%

nH2=2,24/22,4=0,1(mol)

a) PTHH: Zn + 2 HCl -> ZnCl2 + H2

Ta có: nZn=nH2=0,1(mol)

=>a=mZn=0,1.65=6,5(g)

b) nHCl=0,1.2=0,2(mol)

=> CMddHCl=0,2/0,4=0,5(M)

+)Ta có: n_H2=\(\dfrac{2,24}{22,4}\)=0,1(mol)

PTHH: Zn + 2HCl -> ZnCl₂ + H₂

Theo PTHH và đề bài ra ta có: n_Zn=n_H2=0,1(mol)

                                              ⇒n_HCl=2.0,1=0,2(mol)

Khối lượng Zn phản ứng: a=m_Zn=0,1.65=6,5(g)                                                     

+)Ta có: V_ddHCl=400(ml)=0,4(l)

Nồng độ mol của dd HCl tham gia phản ứng: C_{M  ddHCl}=\(\dfrac{0,2}{0,4}\)=0,5(M)

nH2 = 4.48/22.4 = 0.2 (mol) 

Zn + 2HCl => ZnCl2 + H2 

0.2......0.4.....................0.2

mZn = 65*0.2 = 13 (g) 

mHCl = 0.4*36.5 = 14.6 (g) 

C%HCl = 14.6*100%/200 = 7.3%

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

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

3,65% mà bạn nhân sai r

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right);n_{HCl}=0,2.2=0,4\left(mol\right)\\ b,C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)