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

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

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

pứ: Fe + 2HCl -> FeCl₂ + H₂

b. n_Fe = \(\dfrac{5,6}{56}\)= 0,1 mol

Từ pt suy ra được: n_HCl = 2.n_Fe= 0,2 mol

=> m_HCl = 0,2. 36,5 = 7,3 g

c. n_H2 = n_Fe = 0,1 mol

=> V_H2 = 0,1.22,4 = 2,24 (lít)

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

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)

a)

$Fe + 2HCl \to FeCl_2 + H_2$

$n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$

b) $n_{HCl} = 2n_{H_2} = 0,3(mol)$
$\Rightarrow m_{HCl} = 0,3.36,5 = 10,95(gam)$

c) 

Cách 1 : $n_{FeCl_2} = n_{H_2} = 0,15(mol) \Rightarrow m_{FeCl_2} = 0,15.127 = 19,05(gam)$

Cách 2 : Bảo toàn khối lượng, $m_{FeCl_2} = 8,4 + 10,95 - 0,15.2 = 19,05(gam)$

Cíuuuu mn oiii

Câu 1:

\(n_{Fe}=\dfrac{11,2}{56}=0,2(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{H_2}=n_{FeCl_2}=0,2(mol);n_{HCl}=0,4(mol)\\ a,V_{H_2}=0,2.22,4=4,48(l)\\ b,m_{HCl}=0,4.36,5=14,6(g)\\ c,m_{FeCl_2}=0,2.127=25,4(g)\)

Câu 2:

\(n_{Fe}=\dfrac{1,4}{56}=0,025(mol)\)

Theo PT bài 1: \(n_{HCl}=0,05(mol);n_{H_2}=0,025(mol)\\ a,m_{HCl}=0,05.36,5=1,825(g)\\ b,V_{H_2}=0,025.22,4=0,56(l)\)

Câu 3:

\(4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ n_{Al}=\dfrac{2,4.10^{22}}{6.10^{23}}=0,04(mol)\\ \Rightarrow n_{O_2}=0,03(mol);n_{Al_2O_3}=0,02(mol)\\ a,V_{O_2}=0,03.22,4=0,672(l)\Rightarrow V_{kk}=0,672.5=3,36(l)\\ b,m_{Al_2O_3}=0,02.102=2,04(g)\)

Câu 4:

\(S+O_2\xrightarrow{t^o}SO_2\\ a,ĐC:S,O_2\\ HC:SO_2\\ b,n_{O_2}=1,5(mol)\\ \Rightarrow V{O_2}=1,5.22,4=33,6(l)\\ c,d_{S/kk}=\dfrac{32}{29}>1\)

Vậy S nặng > kk

\(a.n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2        0,4          0,2           0,2

\(V_{H_2}=0,2.24,79=4,958l\\ b.m_{HCl}=0,4.36,5=14,6g\\ c.m_{FeCl_2}=0,2.127=25,4g\)

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

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

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

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

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

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

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

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

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

\(b.\) ta có: \(n_{HCl}=2\)

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

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

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

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