PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\), ta được Fe dư.

Theo PT: \(n_{Fe\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

\(\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

\(\Rightarrow m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\)

b, Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

Bạn tham khảo nhé!

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

a) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) \(\Rightarrow\) HCl phản ứng hết, Fe còn dư

\(\Rightarrow n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\) \(\Rightarrow m_{Fe\left(dư\right)}=0,05\cdot56=2,8\left(g\right)\)

b) Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05mol\)

\(\Rightarrow V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\)

\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{22.4}{98}=\dfrac{8}{35}\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

\(2............3\)

\(0.4..........\dfrac{8}{35}\)

\(LTL:\dfrac{0.4}{2}>\dfrac{\dfrac{8}{35}}{3}\Rightarrow Aldư\)

\(m_{Al\left(dư\right)}=\left(0.4-\dfrac{8}{35}\cdot\dfrac{2}{3}\right)\cdot27=6.68\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=\dfrac{8}{35\cdot3}\cdot342=26.05\left(g\right)\)

\(V_{H_2}=\dfrac{8}{35}\cdot22.4=5.12\left(l\right)\)

\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{HCl}=0,5.1=0,5mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1         0,2          0,1          0,1

\(m_{HCl\left(dư\right)}=\left(0,5-0,2\right).36,5=10,95g\\ b)m_{FeCl_2}=0,1.127=12,7g\\ c)V_{H_2}=0,1.22,4=2,24l\\ d)C_{\%HCl\left(dư\right)}=\dfrac{10,95}{200}\cdot100=5,475\%\\ C_{\%HCl\left(pư\right)}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\)

a) n Fe = 5,6/56 = 0,1(mol) ; n HCl = 10,95/36,5 = 0,3(mol)

$Fe + 2HCl \to FeCl_2 + H_2$
Ta thấy : 

n Fe / 1 = 0,1 < n HCl / 2 = 0,3/2 = 0,15 nên HCl dư

Theo PTHH : n HCl pư = 2n Fe = 0,2(mol)

Suy ra: m HCl dư = 10,95 - 0,2.36,5 = 3,65(gam)

b)

Theo PTHH : n FeCl2 = n H2 = n Fe = 0,1(mol)

m FeCl2 = 0,1.127 = 12,7 (gam)

m H2 = 0,1.2 = 0,2(gam)

c)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

n Cu = n H2 = 0,1(mol)

m Cu = 0,1.64 = 6,4(gam)

Fe+2HCl->Fecl2+H2

0,1----0,2----0,1---0,1

n Fe=0,1 mol

n HCl=0,3 mol

=>HCl dư

=>m FeCl2=0,1.127=12,7g

=>VH2=0,1.22,4=2,24l

=>m HCl dư=0,1.36,5=3,65g

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

\(n_{HCl}=\dfrac{10,95}{36,5}=0,3mol\)

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

0,1   < 0,3                          ( mol )

0,1         0,2          0,1       0,1  ( mol )

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

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

\(m_{HCl\left(dư\right)}=\left(0,3-0,2\right).36,5=3,65g\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

LTL: \(0,1< \dfrac{0,3}{2}\rightarrow\) HCl dư

Theo pthh: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=0,1.2=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{HCl\left(dư\right)}=\left(0,3-0,2\right).36,5=3,65\left(g\right)\end{matrix}\right.\)

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 

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

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

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

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

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

Bài 2:

\(a.n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,4}{1}>\dfrac{0,4}{2}\\ \rightarrow Zndư\\ \rightarrow n_{Zn\left(p.ứ\right)}=n_{H_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ \rightarrow n_{Zn\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

b. Sau phản ứng vì Zn dư nên không có phân tử chất nào còn dư.

Bài 1:

\(a.n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,3}{1}< \dfrac{0,8}{2}\\ \rightarrow HCldư\\ n_{H_2}=n_{Mg}=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ n_{HCl\left(dư\right)}=0,8-0,3.2=0,2\left(mol\right)\\ \rightarrow m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)

Fe+2HCl->Fecl₂+H₂

1--------0,2-----0,1----0,1

n Fe=\(\dfrac{5,6}{56}\)=0,1 mol

n HCl=\(\dfrac{36,5}{36,5}\)=1 mol

=>HCl dư :0,8mol

=>m _HCl=0,8.36,5=29,2g

=>m _FeCl2=0,1.127=12,7g

=>V_H2=0,1.22,4=2,24l

a) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\); \(n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{1}{2}\) => Fe hết, HCl dư

PTHH: Fe + 2HCl --> FeCl₂ + H₂

            0,1->0,2----->0,1--->0,1

=> \(m_{HCl\left(dư\right)}=\left(1-0,2\right).36,5=29,2\left(g\right)\)

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

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

a) $Fe + 2HCl \to FeCl_2 + H_2$

b)

n Fe = 0,56/56 = 0,01(mol)

n HCl = 1,095/36,5=  0,03(mol)

Ta thấy :

n Fe / 1 = 0,01 < n HCl /2 = 0,015 => HCl dư

n HCl pư = 2n Fe = 0,02(mol)

=> m HCl dư = 1,095 - 0,02.36,5 = 0,365 gam