a) Coi X là kim loại R hóa trị n

\(2R + 2nHCl \to 2RCl_n + nH_2\\ n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ \Rightarrow n_R = \dfrac{2}{n}n_{H_2} = \dfrac{0,3}{n}(mol)\\ 2R + 2nH_2O \to 2R(OH)_n + nH_2\\ n_{H_2O} = \dfrac{10,8}{18} = 0,6(mol)\\ \Rightarrow n_R = \dfrac{1}{n}n_{H_2O} = \dfrac{0,6}{n}(mol)\\ \)

Suy ra: \(\dfrac{m_1}{m_2} = \dfrac{0,3}{n} : \dfrac{0,6}{n} = \dfrac{1}{2}\)

b)

\(m_2 =2m_1 \\ \Rightarrow C_{M_{HCl\ TN_2}} = 2C_{M_{HCl\ TN_1}} = 0,5.2 = 1M\)

2R+2nHCl→2RCln+nH2nH2=3,3622,4=0,15(mol)⇒nR=2nnH2=0,3n(mol)2R+2nH2O→2R(OH)n+nH2nH2O=10,818=0,6(mol)

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

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

=> \(\left\{{}\begin{matrix}m_{FeCl_2}=0,2.127=25,4\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

Bài 1:

\(PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{NaOH}=\dfrac{6}{40}=0,15\left(mol\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot0,15=0,075\left(mol\right)\\ \Rightarrow m=m_{H_2SO_4}=0,075\cdot98=7,35\left(g\right)\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ \Rightarrow n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V=V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)

Bài 3:

$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$

$Fe+2HCl\to FeCl_2+H_2\uparrow$

Theo PT: $n_{FeCl_2}=n_{H_2}=0,2(mol)$

$\Rightarrow m_{FeCl_2}=0,2.127=25,4(g);V_{H_2}=0,2.22,4=4,48(lít)$

$\Rightarrow m=25,4;V=4,48$

Bài 4:

$CuO+H_2SO_4\to CuSO_4+H_2O$

Theo PT; $n_{H_2SO_4}=n_{CuSO_4}=n_{CuO}=\dfrac{32}{80}=0,4(mol)$

$\Rightarrow m_{H_2SO_4}=0,4.98=39,2(g)$

$m_{CuSO_4}=0,4.160=64(g)$

a,\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:    0,1                           0,05                0,15

\(m_{Al}=0,1.27=2,7\left(g\right)\)

b,\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(mol\right)\)

10.

\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(4Fe+3O_2\underrightarrow{^{^{t^0}}}2Fe_2O_3\)

\(0.3.....0.225....0.15\)

\(V_{O_2}=0.225\cdot22.4=5.04\left(l\right)\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

\(0.15...........0.45\)

\(m_{H_2SO_4}=0.45\cdot98=44.1\left(g\right)\)

11.

\(n_{Fe_2O_3}=\dfrac{48}{160}=0.3\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(0.3...........1.8...........0.6\)

\(m_{FeCl_3}=0.6\cdot162.5=97.5\left(g\right)\)

\(m_{HCl}=1.8\cdot36.5=65.7\left(g\right)\)

Bài 10:

\(a,n_{Fe}=\dfrac{16,8}{56}=0,3(mol)\\ PTHH:4Fe+3O_2\xrightarrow{t^o}2Fe_2O_3\\ Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ \Rightarrow n_{O_2}=\dfrac{3}{4}n_{Fe}=0,225(mol)\\ \Rightarrow V_{O_2}=0,225.22,4=5,04(l)\\ b,n_{H_2SO_4}=3n_{Fe_2O_3}=3.\dfrac{1}{2}n_{Fe}=0,45(mol)\\ \Rightarrow m_{H_2SO_4}=0,45.98=44,1(g)\)

Bài 11:

\(a,n_{Fe_2O_3}=\dfrac{48}{160}=0,3(mol)\\ PTHH:Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ \Rightarrow n_{FeCl_3}=2n_{Fe_2O_3}=0,6(mol)\\ \Rightarrow m_{FeCl_3}=0,6.162,5=97,5(g)\\ b,n_{HCl}=6n_{Fe_2O_3}=1,8(mol)\\ \Rightarrow m_{HCl}=1,8.36,5=65,7(g)\)

Đáp án C

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\\ \left(mol\right)...0,2......\leftarrow...............0,2\\ m_{Zn}=0,2.65=13\left(g\right)\)

PTHH: Zn + 2HCl ^_____>  ZnCl₂ + H₂  (1)

Ta có: theo (1): n_{\(H_2\)(đktc)}=\(\dfrac{4.48}{22.4}\)=0.2 (mol)

theo (1): n_Zn = n_{\(H_2\)}= 0.2(mol)

=> m_Zn = 0.2 . 65 = 13(g)

Vậy giá trị m bằng 13(g)