a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{2}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=2n_{Fe}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,5-0,2=0,3\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)

b, \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

c, \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

d, \(m_{HCl}=0,5.36,5=18,25\left(g\right)\Rightarrow C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)

\(a.n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{HCl}=2n_{Fe}=0,2mol\\ m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95\%\\ b)n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1mol\\ m_{FeCl_2}=0,1.12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(n_{HCl}=0,5.1=0,5\left(mol\right)\)

PTHH :

\(Fr+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,1     0,2           0,1        0,1 

\(\dfrac{0,1}{1}< \dfrac{0,5}{2}\)

--> Tính theo Fe 

HCl dư 

\(a,m_{HCldư}=\left[0,5.\left(0,1.2\right)\right].36,5=10,95\left(g\right)\)

\(b,m_{FeCl_2}=0,1.127=12,7\left(g\right)\)

\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(d,m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(C\%_{HCl}=\dfrac{18,25}{200}.100\%=9,125\%\)

a: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)=n_{FeCl_2}\)

\(\Leftrightarrow m_{FeCl_2}=0.1\cdot127=12.7\left(g\right)\)

b: \(n_{HCl}=2\cdot n_{FeCl_2}=2\cdot0.1=0.2\left(mol\right)\)

\(\Leftrightarrow n_{H_2}=0.1\left(mol\right)\)

\(\Leftrightarrow V_{H_2}=2.24\left(lít\right)\)

ac,nFeCl2=nFe=5,6(mol)

⇒mFeCl2=5,6⋅127=711,2(g)

bSố mol của khí hidro là:        nH2=mH2/MH2=5,6/2=2,8 (mol)

Thể tích khí hidro (ở đktc) là:VH2=nH2x22,9=2,8x22,9=64,12 (lít)

Fe+2HCl->FeCl2+H2

0,6--------------------0,6

H2+CuO-to>Cu+H2O

0,6--------------0,6

n Fe=0,6 mol

=>VH2=0,6.22,4=13,44l

=>m Cu=0,6.64=38,4g

\(n_{Fe}=\dfrac{33,6}{56}=0,6mol\)

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

0,6       1,2          0,6       0,6

\(V_{H_2}=0,6\cdot22,4=13,44l\)

\(CuO+H_2\rightarrow Cu+H_2O\)

0,6       0,6      0,6

\(m_{Cu}=0,6\cdot64=38,4g\)

a. nFe=33.656=0,6(mol)

PTHH : Fe + 2HCl -> FeCl₂ + H₂

            0,6                             0,6 (mol)

VH2=0,6.22,4=13,44(l)

b)

CuO+H2→Cu+H2OCuO+H2→Cu+H2O

0,6       0,6      0,6

mCu=0,6⋅64=38,4g

Em đăng câu hỏi sang môn hóa nha em ơi

a) nFe= 5,6/56=0,1(mol)

nHCl=10,95/36,5=0,3(mol)

PTHH: Fe + 2 HCl -> FeCl2 + H2

Ta có: 0,3/2 > 0,1/1

=> HCl dư, Fe hết, tính theo nFe

-> nH2=nFeCl2=nFe=0,1(mol)

=> V(H2,đktc)=0,1.22,4=2,24(l)

mFeCl2=0,1.127=12,7(g)

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

\(n_{HCl}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)

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

Lập tỉ lệ : 

\(\dfrac{0.2}{1}>\dfrac{0.3}{2}\Rightarrow Fedư\)

Khi đó : 

\(n_{FeCl_2}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.3=0.15\left(mol\right)\)

\(m_{FeCl_2}=0.15\cdot127=19.05\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{2}\\ \Rightarrow HCldư\\ \Rightarrow n_{FeCl_2}=n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{FeCl_2}=127.0,1=12,7\left(g\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

Ta có:n_Fe= 5,6/56=0,1(mol)

n_HCl=10,95/36,5=0,3(mol)

PTHH: Fe + 2 HCl -> FeCl₂ + H₂

Ta có: 0,3/2 > 0,1/1

=> HCl dư, Fe hết, tính theo n_Fe

-> n_H2=n_FeCl2=n_Fe=0,1(mol)

=> V_H2(đktc)=0,1.22,4=2,24(l)

m_FeCl2=0,1.127=12,7(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

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

       0,15<--0,3<-----0,15<--0,15

=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)

=> \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)

b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)