PTHH: \(3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)

Ta có: \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

\(\Rightarrow n_{FeCl_3}=n_{Fe\left(OH\right)_3}=\dfrac{1}{12}\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_3}=\dfrac{1}{12}\cdot162,5\approx13,54\left(g\right)\\m_{Fe\left(OH\right)_3}=\dfrac{1}{12}\cdot107\approx8,92\left(g\right)\end{matrix}\right.\)

nNaOH =  m/M = 10/(23 +16 + 1) = 0,25 (mol)

Ta có PTHH:   3NaOH   +   FeCl3   ------>   Fe(OH)3   +   3NaCl

Theo PT:        3           -          1          -               1                (mol)

BC:           0.25           -        0.083      -           0.083              (mol)

Suy ra:    mFeCl3 = n x M = 0.083 x (56 + 35,5 x 3) = 13,4875 (g)

mFe(OH)3 = n x M = 0,083 x (56+17 x 3) = 8,881 (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

Fe + 2HCl -> FeCl2 + H2 

giải nhanh hộ e vs ạ e cần gấp ạ 

\(HCl+NaOH\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaCl}=n_{NaOH}=1,5.0,1=0,15\left(mol\right)\\ a,m_{HCl}=0,15.36,5=5,475\left(g\right)\\ b,m_{NaCl}=58,5.0,15=8,775\left(g\right)\)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{^{t^0}}}FeCl_3\)

\(0.1.......0.15..........0.1\)

\(V_{Cl_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{FeCl_3}=0.1\cdot162.5=16.25\left(g\right)\)

a.3.36l

b.13.25g

a) nNaOH= 6/40=0,15(mol)

nFeCl3=32,5/162,5= 0,2(mol)

PTHH: 3 NaOH + FeCl3 -> Fe(OH)3 + 3 NaCl

0,15________0,05____0,05________0,15(mol)

Ta có: 0,2/1 > 0,15/3

=> NaOH hết, FeCl3 dư

=> nFeCl3(dư)= 0,2-0,05=0,15(mol)

=> mFeCl3= 162,5.0,15=24,375(g)

b)m(kết tủa)= mFe(OH)3= 0,05.107= 5,35(g)

cảm ơn nhiều ạ

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

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

\(m_{FeCl_2}=0,2.127=25,4g\\ b.m_{HCl}=0,4.36,5=14,6g\)

a) PTHH: FeCl₃ + 3KOH → Fe(OH)₃ + 3KCl

b) Theo ĐLBTKL ta có:

 \(m_{FeCl_3}+m_{KOH}=m_{Fe\left(OH\right)_3}+m_{KCl}\)

\(\Leftrightarrow m_{FeCl_3}=m_{Fe\left(OH\right)_3}+m_{KCl}-m_{KOH}=2,14+4,47-3,36=3,25\left(g\right)\)

a. \(Fe+2HCl->FeCl_2+H_2\)

b. Áp dụng định luật bảo toàn khối lượng:

\(m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\)

\(8,4+m_{HCl}=19,05+0,3\)

\(8,4+m_{HCl}=19,35\)

         \(m_{HCl}=19,35-8,4=10,95\left(g\right)\)