\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.2......0.4..........0.2...........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(PTPU:Fe+2HCl\rightarrow FeCl_2+H_2\)

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

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

\(b.\) ta có: \(n_{HCl}=2\)

\(\Rightarrow n_{Fe}=0,2.2=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(c.n_{FeCl_2}=n_{Fe}=0,2mol\)

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

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,2--->0,4---->0,2----->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 =14,6 (g)

c) m_FeCl2 = 0,2.127 = 25,4 (g)

Câu 1:

PTHH: Fe + 2HCl ===> FeCl₂ + H₂

a/ n_Fe = 11,2 / 56 = 0,2 mol

=> n_H2 = 0,2 mol

=> V_H2(đktc) = 0,2 x 22,4 = 4,48 lít

b/ => n_HCl = 0,2 x 2 = 0,4 mol

=> m_HCl = 0,4 x 36,5 = 14,6 gam

c/ => n_FeCl2 = 0,2 mol

=> m_FeCl2 = 0,2 x 127 = 25,4 gam

Câu 3/

a/ Chất tham gia: S, O₂

Chất tạo thành: SO₂

Đơn chất: S, O₂ vì những chất này chỉ do 1 nguyên tố tạo nên

Hợp chất: SO₂ vì chất này do 2 nguyên tố S và O tạo tên

b/ PTHH: S + O₂ =(nhiệt)==> SO₂

=> n_O2 = 1,5 mol

=> V_O2(đktc) = 1,5 x 22,4 = 33,6 lít

c/ Khí sunfuro nặng hơn không khí

m

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

a) PTHH: Fe + 2HCl --> FeCl₂ + H₂

_______0,2---->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48(l)

b) m_HCl = 0,4.36,5 = 14,6(g)

c) m_FeCl2 = 0,2.127 = 25,4 (g)

\(n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ 0,03....0,06.....0,03.......0,03\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,03.22,4=0,672\left(l\right)\\ b,m_{HCl}=0,06.36,5=2,19\left(g\right)\\ c,m_{FeCl_2}=127.0,03=3,81\left(g\right)\)

\(Fe+2HCl-->FeCl_2+H_2\)

0,2___0,4__________0,2____0,2

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

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

b)=> \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

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

Fe + 2HCl → FeCl₂ + H₂

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

a) Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)

b) Theo PT: \(n_{HCl}=2n_{Fe}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,4\times36,5=14,6\left(g\right)\)

c) Theo PT: \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,2\times127=25,4\left(g\right)\)

Fe+2HCl---------> FeCl2 + H2

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

a) Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)

=> V _H2 = 0,2. 22,4 = 4,48(l)

b) Theo PT: \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\)

=> m _HCl = 0,4. 36,5 = 14,6 (g)

c) Theo PT: \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

=> m _FeCl2 = 0,2. 127 = 25,4 (g)

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)

________0,2____0,4______ 0,2_____0,2

a, Ta có:

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

\(\Rightarrow V_{H2}=0,2.22,4=4,48\left(l\right)\)

b,

\(m_{HCl}=0,4.\left(1+35,5\right)=14,6\left(g\right)\)

c,

\(m_{FeCl2}=0,2.\left(56+35,5.2\right)=25,4\left(g\right)\)

\(a,n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\)

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow n_{H_2}=n_{Fe}=0,03\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,03\cdot22,4=0,672\left(l\right)\\ b,n_{HCl}=2n_{Fe}=0,06\left(mol\right)\\ \Rightarrow m_{HCl}=0,06\cdot36,5=2,19\left(g\right)\\ c,n_{FeCl_2}=n_{Fe}=0,03\left(mol\right)\\ \Rightarrow m_{FeCl_2}=0,03\cdot127=3,81\left(g\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ a.n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\\ n_{H_2}=n_{Fe}=0,03\left(mol\right)\\ \Rightarrow V_{H_2}=0,03.22,4=0,672\left(l\right)\\ b.n_{HCl}=2n_{Fe}=0,06\left(mol\right)\\ m_{HCl}=0,06.36,5=2,19\left(g\right)\\ c.n_{FeCl_2}=n_{Fe}=0,03\left(mol\right)\\ m_{FeCl_2}=0,03.127=3,81\left(g\right)\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)