1:

a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

______0,2------>0,2------------------->0,2_____(mol)

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

b) \(V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)\)

2:

a) 

\(n_{HCl}=2.0,2=0,4\left(mol\right)\)

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

______0,2<------0,4------------------>0,2______(mol)

=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)

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

\(19.\\ a)n_{Mg}=\dfrac{2,4}{24}=0,1mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{H_2}=n_{H_2SO_4}=n_{Mg}=0,1mol\\ V_{H_2}=0,1.22,4=2,24l\\ b)m_{ddH_2SO_4}=\dfrac{0,1.98}{10}\cdot100=98g\)

\(20.\\ a)n_{NaOH}=0,2.1=0,2mol\\ 2NaOH+H_2SO_4\rightarrow NaSO_4+2H_2O\)

0,2                0,1                0,1            

\(m_{Na_2SO_4}=0,1.142=14,2g\\ b)m_{ddH_2SO_4}=\dfrac{0,1.98}{5}\cdot100=196g\)

Câu 1 :

n Mg = 4,8/24 =0,2(mol)

n H2 = 3,36/22,4 = 0,15(mol)

Y gồm 0,2 mol Mg và O

Bảo toàn electron :

2n Mg = 2n O + 2n H2

<=> n O = 0,2 - 0,15 = 0,05(mol)

\(2H^+ + O^{2-} \to H_2O\\ 2H^+ + 2e \to H_2\)

Ta có :

n H+ = 2n O + 2n H2 = 0,15.2 + 0,05.2 = 0,4(mol)

=> n H2SO4 = 1/2 nH+ = 0,2(mol)

=> V dd H2SO4 = 0,2/1 = 0,2(lít)

Câu 2  :

Oleum : H2SO4.nSO3

n NaOH = 0,2.0,15 = 0,03(mol)

2NaOH + H2SO4 $\to$ Na2SO4 + 2H2O

n H2SO4 = 1/2 n NaOH = 0,015(mol)

=> trong 200 ml dung dịch X chứa 0,015.2 = 0,03(mol) H2SO4

H2SO4.nSO3 + nH2O $\to$ (n + 1)H2SO4

Theo PTHH :

\(n_{oleum} = \dfrac{n_{H_2SO_4}}{n + 1}\\ \Rightarrow 0,015 = \dfrac{0,03}{n + 1}\\ \Rightarrow n = 1\)

Vậy oleum cần tìm là H2SO4.SO3

$n_{Mg} = \dfrac{4,8}{24} = 0,2(mol) ; n_{HCl} = 0,5(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
Ta thấy : 

$n_{Mg} : 1 < n_{HCl} : 2$ nên HCl dư

$n_{H_2} = n_{Mg} = 0,2(mol) \Rightarrow V_{H_2} = 0,2.22,4 = 4,48(lít)$

\(a)H_2SO_4+Ba\left(NO_3\right)_2\rightarrow BaSO_4+2HNO_3\\ n_{H_2SO_4}=0,2.1=0,2l\\ n_{Ba\left(NO_3\right)_2}=n_{H_2SO_4}=0,2mol\\ m_{ddBa\left(NO_3\right)_2}=\dfrac{0,2.261}{20}\cdot100=261g\\ V_{ddBa\left(NO_3\right)_2}=\dfrac{261}{1,22}\approx213,9ml\\ c)n_{HNO_3}=0,2.4=0,4mol\\ C_{M_{HNO_3}}=\dfrac{0,4}{0,2+0,2139}\approx0,97M\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\), ta được HCl dư.

Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{H_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,4\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\)

b, Dung dịch a gồm HCl dư và FeCl₂.

PT: \(HCl+NaOH\rightarrow NaCl+H_2O\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_{2\downarrow}+2NaCl\)

Theo PT: \(n_{NaOH}=n_{HCl\left(dư\right)}+2n_{FeCl_2}=0,5\left(mol\right)\)

\(\Rightarrow a=C_{M_{NaOH}}=\dfrac{0,5}{0,2}=2,5M\)

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

Bạn tham khảo nhé!

a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_{HCl}=0,2.1=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

Xét tỉ lệ:  \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => Zn dư, HCl hết

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

__________0,2-------------->0,1

=> V_H2 = 0,1.22,4 = 2,24(l)

b)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

______0,1->0,05

=> m_O2 = 0,05.22,4 = 1,12 (l)

nMg = 4,8:24= 0,2(mol) 
pthh : Mg+H2SO4 --> MgSO4 +H2
          0,2----------------------------> 0,2(mol) 
=> VH2= 0,2.22,4 =4,48(l) 
b) H2+ CuO -t--> Cu +H2O 
    0,2------------ >0,2(MOL)
=> mCu = 0,2.64=12,8 (g)