\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: Zn + 2HCl ---> FeCl₂ + H₂

          0,3<---------------0,3<----0,3

=> \(\left\{{}\begin{matrix}m=0,3.65=19,5\left(g\right)\\m_{muối}=0,3.136=40,8\left(g\right)\\V_{ddHCl}:thiếu.C_M\end{matrix}\right.\)

\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O
LTL: \(0,2>\dfrac{0,3}{3}\) => Fe₂O₃ dư

Theo pthh: n_{Fe2O3 (pư)} = \(\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

                  n_Fe = \(\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)

=> m_{chất rắn} = 0,1.160 + 0,2.56 = 27,2 (g)

Zn+2HCl->ZnCl2+H2

0,3---0,6----0,3----0,3

n H2=0,3 mol

=>m Zn=0,3.65=19,5g

=>m muối=0,3.136=40,8g

c) thiếu đề

d)

Fe2O3+3H2-to>2Fe+3H2O

               0,3------0,2

n Fe2O3=0,2 mol

=>Fe2O3 dư

=>m cr=0,2.56+0,1.160=27,2g

\(n_{CaO}=\dfrac{8}{40}=0,2\left(mol\right)\)

Pt : \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2|\)

         1         2                1            1

        0,2                         0,2         0,2

a) \(n_{Ca\left(OH\right)2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{Ca\left(OH\right)2}=0,2.74=14,8\left(g\right)\)

b) \(m_{ddspu}=8+100-\left(0,2.2\right)=107,6\left(g\right)\)

\(C_{ddNaOH}=\dfrac{14,8.100}{107,6}=13,75\)⁰/₀

c)Pt :  \(2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O|\)

              2              1                  1             2

            0,4             0,2

\(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

\(V_{ddHCl}=\dfrac{0,4}{0,5}=0,8\left(l\right)\)

                       = 800ml

 Chúc bạn học tốt

a. PTHH : Fe + 2HCl -> FeCl₂ + H₂ 

b. Ta có PTHH : Fe + 2HCl -> FeCl₂ + H₂ 

Theo pt             1mol  2mol

Theo đề          0,1mol->0,2mol

Khối lượng HCl đã phản ứng :

\(m_{HCl}\) = \(n_{HCl}\) . \(M_{HCl}\)

=> \(m_{HCl}\) = 0,2 . 36,5

=>\(m_{HCl}\) = 7,3 (g)

a, Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

a--->2a------------------>a

2Al + 6HCl ---> 2AlCl₃ + 3H₂

b---->3b-------------------->1,5b

=> \(\left\{{}\begin{matrix}56a+27b=16,6\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow a=b=0,2\left(mol\right)\)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) \(C\%_{HCl}=\dfrac{\left(0,2.2+0,2.3\right).36,5}{300}.100\%=12,167\%\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) 
gọi nFe : a , nAl: b (a,b>0)  => 56a + 27b = 16,6 (g) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
           a                                     a
         \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\) 
           b                                     \(\dfrac{3b}{2}\)
          
=> \(a+\dfrac{3b}{2}=0,5\) 
ta có hệ pt 
\(\left\{{}\begin{matrix}56a+27b=16,6\\a+\dfrac{3b}{2}=0,5\end{matrix}\right.\) 
=> a= 0,2 , b = 0,2 
\(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=16,6-11,2=5,4\left(g\right)\end{matrix}\right.\) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,2     0,4 
        \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,2    0,6 
=> \(m_{HCl}=\left(0,4+0,6\right).36,5=36,5\left(g\right)\) 
=> \(C\%=\dfrac{36,5}{200}.100\%=18,25\%\)

a)

\(2Al + 6HCl \to 2AlCl_3 + 3H_2 \)

b)

Tỉ lệ số nguyên tử Al : số phân tử HCl : số phân tử AlCl₃ : số phân tử H₂ là 2 : 6 : 2  : 3

c)

Ta có : \(n_{Al} = \dfrac{13,5}{27} = 0,5(mol)\)

Theo PTHH : \(n_{H_2} = \dfrac{3}{2}n_{Al} = 0,75(mol)\\ \Rightarrow V_{H_2} = 0,75.22,4 = 16,8(lít)\)

a) Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,1--------------->0,1---->0,1

=> m_FeCl2  = 0,1.127 = 12,7(g)

c) V_H2 = 0,1.22,4 = 2,24(l)

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

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{24,6375}{36,5}=0,675\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

LTL: \(\dfrac{0,2}{2}< \dfrac{0,675}{3}\rightarrow\) HCl dư

Theo pthh: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=3n_{Al}=3.0,2=0,6\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\m_{HCl\left(dư\right)}=\left(0,675-0,5\right).36,5=2,7375\left(g\right)\\V_{H_2}=0,3.22,4=6,72\left(l\right)\end{matrix}\right.\)

2Al+6HCl->2AlCl3+3H2

0,2-----0,6-----0,2-----0,3

n Al=0,2 mol

n HCl=0,675 mol

=>Hcl dư

=>m HCl=0,075.36,5=2,7375g

b)m AlCl3=0,2.133,5=26,7g

c) VH2=0,3.22,4=6,72l

a) $n_{H_2SO_4} = \dfrac{44,1}{98} = 0,45(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = 0,3(mol)$
$m_{Al} = 0,3.27 = 8,1(gam)$

b) $n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$\Rightarrow V_{H_2} = 0,45.22,4 =1 0,08(lít)$

c)

Cách 1 : $n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,15(mol)$
$\Rightarrow m_{Al_2(SO_4)_3} = 0,15.342 = 51,3(gam)$

Cách 2 : Bảo toàn khối lượng, $m_{Al_2(SO_4)_3} = 8,1 + 44,1 - 0,45.2 = 51,3(gam)$

Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

_____0,05__0,1____________0,05 (mol)

b, m_Fe = 0,05.56 = 2,8 (g)

c, m_HCl = 0,1.36,5 = 3,65 (g)

\(\Rightarrow m_{ddHCl}=\dfrac{3,65}{10\%}=36,5\left(g\right)\)

Bạn tham khảo nhé!

\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

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

\(0.05......0.1...................0.05\)

\(m_{Fe}=0.05\cdot56=2.8\left(g\right)\)

\(m_{dd_{HCl}}=\dfrac{0.1\cdot36.5\cdot100}{10}=36.5\left(g\right)\)

Ta có: \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

\(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\uparrow\)

               0,4 <---  0,6    ----------->      0,2    -->   0,6

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\\V_{H_2}=0,6.22,4=13,44\left(lít\right)\end{matrix}\right.\)