Ta có : n_H2 = V_H2 / 22,4 = 1,12 / 22,4 = 0,05 mol ( ĐKTC )

Theo PTHH : n_R = n_H2 = 0,05 ( mol )

Mà n_R = m_R / M_R = 4,8 / M_R ( mol )

-> 4,8 / M_R = 0,05

-> M_R = 96

-> R là Cm

\(a,PTHH:R+2HCl\rightarrow RCl_2+H_2\)

Theo phương trình trên:
\(n_R=n_{RCl_2}\)

\(\Leftrightarrow\frac{13}{R}=\frac{27,2}{R+71}\)

\(\Leftrightarrow27,2R=13\left(R+71\right)\)

\(\Leftrightarrow R=65\).

Vậy \(R\) là \(Zn\)

\(b,2R+2n_{HCl}\rightarrow2RCln+n_{H_2}\)

\(n_{H_2}=\frac{4,704}{22,4}=0,21\left(mol\right)\)
\(\Rightarrow n_R=\frac{0,42}{n}mol\)
\(M_R=\frac{3,78n}{0,42}=9n\)

\(n=3\Rightarrow M_R=27\)

Vậy \(R\) là \(Al\)

\(n_{H_2O}=\dfrac{1.8}{18}=0.1\left(mol\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)

\(0.1.........0.25......0.2..........0.1\)

\(V_{C_2H_2}=2.24\left(l\right)\)

\(V_{O_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(V_{CO_2}=0.2\cdot22.4=4.48\left(l\right)\)

B1 :

PTHH : Zn + 2HCl ->(t*) ZnCl₂ + H₂

Theo đề bài ta có : n_Zn = \(\dfrac{m}{M}=\dfrac{6.5}{65}=0,1\left(mol\right)\)

Theo PTHH ta có \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

=> \(V_{H_2\left(\text{đ}ktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\)

Theo PTHH ta có \(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)

=> \(m_{ZnCl_2}=n.M=0,1.136=13,6\left(g\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

           0,2       0,3                0,1             0,3

\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

b) V = 100ml = 0,1l 

\(C_{MH2SO4}=\dfrac{0,3}{0,1}=3\left(M\right)\)

  \(m_{muối}=m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)

 Chúc bạn học tốt

a,2Al + 3H₂SO₄  → Al₂(SO₄)₃ + 3H₂ 

n_Al = 5,4 : 27 = 0,2mol

n_{H\(_2\)}=0,2.3:2 =0,3mol

V_{H\(_2\)} = 0,3.22,4 =6,72 l

b. n_{H\(_2\)SO\(_4\)} = 0,2.3:2=0,3mol

C_{M H\(_2\)SO\(_4\)} = 0,3:0,1 =3M

n_{Al\(_2\)(SO\(_4\))\(_3\)} = 0,2:2 =0,1mol

m_{\(Al_2 (SO_4 ) _3\)} =0,1. 342 =34,2g

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)

\(0.1....0.075.....0.05\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(m_{Al_2O_3}=0.05\cdot102=5.1\left(g\right)\)

ta có: \(n_{Al}=0.1\left(mol\right)\)

PTHH

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

4         3

0.1      x

\(=>x=0.075=n_{O_2}\)

\(=>V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ b.n_{H_2}=n_{Fe}=\dfrac{10}{56}=\dfrac{5}{28}\left(mol\right)\\ \Rightarrow V_{H_2}=\dfrac{5}{28}.22,4=4\left(l\right)\)

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

\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)

\(0.1.......0.075.....0.05\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(m_{Fe_2O_3}=0.05\cdot160=8\left(g\right)\)

PTHH: \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\)

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

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{Fe_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)