1.\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4mol\)

\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)

\(\Rightarrow n_{H_2}=n_{H_2O}=0,4mol\)

\(m_{H_2}=n.M=0,4.2=0,8g\)

\(m_{H_2O}=n.M=0,4.18=7,2g\)

Định luật BTKL:

\(m_{Fe_xO_y}+m_{H_2}=m_{Fe}+m_{H_2O}\)

\(\Rightarrow m_{Fe}=16,8g\)

\(n_{O\left(trong.oxit\right)}=n_{H_2O}=n_{H_2}=0,4mol\)

\(n_{Fe\left(trong.oxit\right)}=n_{Fe}=\dfrac{16,8}{56}=0,3mol\)

\(x:y=0,3:0,4=3:4\)

Vậy \(CTHH:Fe_3O_4\)

2.

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25mol\)

\(4M+O_2\rightarrow\left(t^o\right)2M_2O\)

  1      0,25              0,5        ( mol )

\(m_{O_2}=n.M=0,25.32=8g\)

Định luật BTKL:

\(m_M+m_{O_2}=m_{M_2O}\)

\(\Rightarrow m_M=39g\)

\(M_M=\dfrac{m}{n}=\dfrac{39}{1}=39\) ( g/mol )

\(\Rightarrow M:Kali\left(K\right)\)

\(n_{H_2S}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{NaOH}=0,25.2=0,5\left(mol\right)\)

Xét \(T=\dfrac{0,5}{0,4}=1,25\) => Tạo cả 2 muối Na₂S và NaHS

PTHH: 

2NaOH + H₂S ---> Na₂S + 2H₂O

0,5------->0,25------>0,25

Na₂S + H₂S ---> 2NaHS 

0,15<--0,15----->0,3

=> m_muối = 0,3.56 + (0,25 - 0,15).78 = 24,6 (g)

Chọn A

Đáp án A

n_X = 0,4 (mol)

n_↓ = n_PbS = 0,1 (mol)

Pb(NO₃)₂ + H₂S →PbS + 2HNO₃

a) 

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

             0,4-->0,2------->0,4

=> \(m_{H_2O}=0,4.18=7,2\left(g\right)\)

b) V_O2 = 0,2.22,4 = 4,48 (l)

c)

C1: m_O2 = 0,2.32 = 6,4 (g)

C2:

Theo ĐLBTKL: \(m_{O_2}=m_{H_2O}-m_{H_2}=7,2-0,4.2=6,4\left(g\right)\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: Zn + H₂SO₄ --> ZnSO₄ + H₂

=> n_Zn = 0,4 (mol)

=> m_Zn = 0,4.65 = 26(g)

Zn+H₂SO₄->ZnSO₄+H₂

0,4------------------------0,4 mol

n _H2=\(\dfrac{8,96}{22,4}\)=0,4 mol

=>a=m _Zn=0,4.65=26g

\(n_{H2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(n_{H2}=n_{H2SO4}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{ddH2SO4}=\dfrac{0,4.98}{200}.100\%=19,6\%\)

1)

a) \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

b) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

c) \(n_{H_2O}=\dfrac{15.10^{23}}{6.10^{23}}=2,5\left(mol\right)\)

2)

a) \(n_A=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) => M_A = \(\dfrac{3}{0,1}=30\left(g/mol\right)\)

b) \(d_{A/O_2}=\dfrac{30}{32}=0,9375\)