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

\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ b.n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2}=0,2.22,4=4,48\left(l\right)\\ c.n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ m_{MgCl_2}=0,2.95=19\left(g\right)\)

a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Theo PT: \(n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,5-0,2=0,3\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)

b, \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

c, \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

d, \(m_{HCl}=0,5.36,5=18,25\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)

\(a.n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{HCl}=2n_{Fe}=0,2mol\\ m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95\%\\ b)n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1mol\\ m_{FeCl_2}=0,1.12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

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

PTHH :

\(Fr+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,1     0,2           0,1        0,1 

\(\dfrac{0,1}{1}< \dfrac{0,5}{2}\)

--> Tính theo Fe 

HCl dư 

\(a,m_{HCldư}=\left[0,5.\left(0,1.2\right)\right].36,5=10,95\left(g\right)\)

\(b,m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

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

\(d,m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)

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

PTHH: Mg + 2HCl --> MgCl₂ + H₂

_____0,2------------------------>0,2

=> V_H2 = 0,2.24,79 = 4,958 (l)

\(n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{AlCl_3} = n_{Al} = 0,4(mol)\\ m_{AlCl_3} = 0,4.133,5 = 53,4(gam)\\ n_{HCl} =3 n_{Al} = 1,2(mol)\\ C\%_{HCl}= \dfrac{1,2.36,5}{200}.100\% = 21,9\%\)

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

Đề bài phải là thể tích CO₂ bạn nhé!

a, PT: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

Ta có: \(n_{CaCO_3}=\dfrac{4}{100}=0,04\left(mol\right)\)

\(m_{HCl}=\dfrac{14,6.25}{100}=3,65\left(g\right)\Rightarrow n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,1}{2}\), ta được HCl dư.

Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,04\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,04.22,4=0,896\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{CaCO_3}=0,08\left(mol\right)\\n_{CaCl_2}=n_{CaCO_3}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,02\left(mol\right)\Rightarrow m_{HCl\left(dư\right)}=0,02.36,5=0,73\left(g\right)\)

\(m_{CaCl_2}=0,04.111=4,44\left(g\right)\)

Bạn tham khảo nhé!

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

Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{4}{100}=0,04\left(mol\right)\\n_{HCl}=\dfrac{14,6\cdot25\%}{36,5}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,1}{2}\) \(\Rightarrow\) HCl còn dư, CaCO₃ p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=n_{CaCl_2}=0,04\left(mol\right)\\n_{HCl\left(dư\right)}=0,02\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,04\cdot22,4=0,896\left(l\right)\\m_{CaCl_2}=0,04\cdot111=4,44\left(g\right)\\m_{HCl\left(dư\right)}=0,02\cdot36,5=0,73\left(g\right)\end{matrix}\right.\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2\cdot36,5}{200}\cdot100\%=3,65\%\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)