\(n_{H_2S}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(n_{SO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Gọi số mol Mg là x (mol)

Mg⁰ - 2e --> Mg⁺²

x--->2x

S⁺⁶ + 8e --> S^-2

         1,2<--0,15

S⁺⁶ + 2e --> S⁺⁴

         0,4<--0,2

Bảo toàn e: 2x = 1,6 (mol)

=> x = 0,8 (mol)

=> m = 0,8.24 = 19,2 (g)

bảo toàn Mg: n_MgSO4 = 0,8 (mol)

=> m₁ = 0,8.120 = 96 (g)

nH2=13,14:22,4=0,6 mol

PTHH: 2Al+6HCl=>2Al2Cl3+3H2

             0,4<-1,2<----0,4<-----0,6

=> Al=0,4.27=10,8g

CMHCL=1,2:0,4=3M

CM Al2Cl3=0,4:0,4=1M

bài 2: nH2=0,2mol

PTHH: 2A+xH2SO4=> A2(SO4)x+xH2

             0,4:x<---------------------------0,2

ta có PT: \(\frac{13}{A}=\frac{0,4}{x}\)<=> 13x=0,4A

=> A=32,5x

ta lập bảng xét

x=1=> A=32,5   loiaj

x=2=> A=65    nhận

x=3=> A=97,5    loại

=> A là kẽm (Zn) 

a) PTHH: Mg +2 HCl -> MgCl2 + H2

 nH2=11,2/22,4=0,5(mol)

=> nMg=nMgCl2=nH2=0,5(mol)

m=mMg=0,5.24=12(g)

b) mMgCl2=0,5.95=47,5(g)

mddMgCl2= mMg + mddHCl - mH2= 12+100-0,5.2= 111(g)

=>C%ddMgCl2= (47,5/111).100=42,793%

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

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

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

b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)

c, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

Bạn tham khảo nhé!

PTHH: Fe + CuSO₄ → FeSO₄ + Cu

Mol:      x         x               x           x

Theo ĐLBTKL,ta có: 56x+160x = 152x + 64x

                            ⇔  160x - 152x = 64x - 56x = m+16-m=16

                            ⇔   8x = 16

                            ⇔    x=2

⇒ m=m_Fe = 56.2 = 112 (g)

Vậy m=112 g

\(Fe+CuSO_4 \to FeSO_4+Cu\\ n_{Fe}=a(mol)\\ n_{Cu}=a(mol)\\ m+16=64a\\ \to 1,6=64a-56a\\ a=0,2(mol)\\ m_{Fe}=0,2.56=11,2(g)\)

\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(\Rightarrow n_{HCl}=2\cdot n_{H_2}=2\cdot0.3=0.6\left(mol\right)\)

BTKL : 

\(m_{Muối}=m_{hh}+m_{HCl}-m_{H_2}=18.6+0.6\cdot36.5-0.3\cdot2=39.9\left(g\right)\)

\(n_{Cl_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PTHH: KClO₃ + 6HCl --> KCl + 3Cl₂ + 3H₂O

              0,15<-------------------0,45

=> \(H=\dfrac{0,15.122,5}{24,5}.100\%=75\%\)

- PT: a, \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\) (1)

\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\) (2) 

- Ta có: \(n_{HCl\left(1\right)}=n_{HCl\left(2\right)}=0,2.2=0,4\left(mol\right)\)

Theo PT (1): \(n_{Cl_2}=\dfrac{5}{16}n_{HCl\left(1\right)}=0,125\left(mol\right)\Rightarrow V_1=0,125.22,4=2,8\left(l\right)\)

(2): \(n_{Cl_2\left(2\right)}=\dfrac{1}{4}n_{HCl\left(2\right)}=0,1\left(mol\right)\Rightarrow V_2=0,1.22,4=2,24\left(l\right)\)