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Bài 1:

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

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

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

Theo PT: \(n_{Mg\left(pư\right)}=n_{MgCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)

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

\(\Rightarrow m_{Mg\left(dư\right)}=0,05.24=1,2\left(g\right)\)

\(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)

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

Bài 2:

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

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

\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,15}{3}\) , ta được Al dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Al\left(pư\right)}=\dfrac{2}{3}n_{H_2SO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,05\left(mol\right)\\n_{H_2}=n_{H_2SO_4}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{Al\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

Bài 3:

PT: \(2M+6HCl\rightarrow2MCl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{4,704}{22,4}=0,21\left(mol\right)\)

Theo PT: \(n_M=\dfrac{2}{3}n_{H_2}=0,14\left(mol\right)\)

\(\Rightarrow M_M=\dfrac{3,78}{0,14}=27\left(g/mol\right)\)

Vậy: M là nhôm (Al).

Bài 4:

PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,2\left(mol\right)\)

PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{4}>\dfrac{0,2}{5}\) , ta được P dư.

Theo PT: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,08\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,08.142=11,36\left(g\right)\)

Bạn tham khảo nhé!

a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1                         0,1      0,15   ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{AlCl_3}=0,1.133,5=13,35g\)

d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

                  0,15              0,1              ( mol )

\(m_{Fe}=0,1.56=5,6g\)

a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,4}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

           0,1--->0,3------>0,1---->0,15

=> m_HCl = (0,4 - 0,3).36,5 = 3,65 (g)

b) V_H2 = 0,15.22,4 = 3,36 (l)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)

\(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{14,6}{36,5}=0,4mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1  <   0,4                                 ( mol )

0,1        0,3                         0,15 ( mol )

a. Chất còn dư là HCl

\(m_{HCl}=n_{HCl}.M_{HCl}=\left(0,4-0,3\right).36,5=3,65g\)

\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36l\)

a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

b) Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\\n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,1}{2}\) \(\Rightarrow\) HCl phản ứng hết, Zn còn dư

\(\Rightarrow n_{Zn\left(dư\right)}=0,5-0,05=0,45\left(mol\right)\) \(\Rightarrow m_{Zn\left(dư\right)}=0,45\cdot65=29,25\left(g\right)\)

c+d) Theo PTHH: \(n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05mol\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,05\cdot136=6,8\left(g\right)\\V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\end{matrix}\right.\)

\(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

\(n_{HCl}=2.0,2=0,4\left(mol\right)\)

PTHH :

                 \(CuO+2HCl\rightarrow CuCl_2+H_2\uparrow\)

trc p/ư:      0,15      0,4 

p/ư :          0,15      0,3       0,15          0,15 

sau p/ư :   0        0,1         0,15        0,15 

--> sau p/ư : HCl dư 

\(a,m_{CuCl_2}=0,15.135=20,25\left(g\right)\)

\(b,C_{M\left(CuCl_2\right)}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

\(a)n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2\\ \dfrac{0,15}{1}< \dfrac{0,4}{2}\Rightarrow HCl.dư\\ n_{CuCl_2}=n_{CuO}=n_{H_2}=0,15mol\\ m_{CuCl_2}=0,15.135=20,25\left(g\right)\\ b)C_{MCuCl_2}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\ n_{HCl\left(pư\right)}=0,15.2=0,3\left(mol\right)\\ n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\\ C_{MHCl\left(dư\right)}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2.......0.4.......................0.2\)

\(m_{Zn}=0.2\cdot65=13\left(g\right)\)

\(C\%_{HCl}=\dfrac{0.4\cdot36.5}{200}\cdot100\%=7.3\%\)

\(n_{CuO}=\dfrac{24}{80}=0.3\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(1..........1\)

\(0.3.........0.2\)

\(LTL:\dfrac{0.3}{1}>\dfrac{0.2}{1}\Rightarrow CuOdư\)

\(m_{CuO\left(dư\right)}=\left(0.3-0.2\right)\cdot64=6.4\left(g\right)\)

\(n_{HCl}=\dfrac{m}{M}=\dfrac{36,5}{36,5}=1\left(mol\right)\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{65}{65}=1\left(mol\right)\)

                     Zn  +   2HCl   →  ZnCl₂ + H₂

Theo đề:    1 mol     2 mol

Pư:             1 mol    1 mol

Tỉ lệ:           \(\dfrac{1}{1}>\dfrac{1}{2}\)             \(\Rightarrow\) Zn là chất thừa, tính theo HCl

   Zn  +   2HCl   →  ZnCl₂ + H₂

                1                   →    1             ( mol )

\(V_{H_2}=n.22,4=1.22,4=22,4\left(l\right)\)

n_Zn = m_Zn:M_Zn = 65:65 = 1mol

n_HCl = m_HCl:M_HCl = 36,5:36,5 = 1mol

          Zn + 2HCl -> ZnCl₂ + H₂

Theo PTHH:1mol 2mol            1mol

Theo ĐB:  1mol 1mol            1mol

a, Vì 1mol/1mol > 1mol/2mol => HCl sẽ hết trước => Tính theo Zn

b, m_H2 = n_H2.22,4 = 2.22,4 = 44,8ml

Bài 1:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right);n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

TheoPT: 4mol....3mol

Theo đề:0,1.......0,15 mol

Tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,15}{3}\)

=> \(O_2\) dư

Theo PTHH : \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\Rightarrow m_{Al}=27.0,05=1,35\left(g\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}.0,1=0,075\left(mol\right)\)

=> \(n_{O_2dư}=0,15-0,075=0,075\left(mol\right)\)

\(\Rightarrow m_{O_2dư}=0,075.32=2,4\left(g\right)\)

Bài 2 :

Tương tự

Tìm chất dư, rồi tính theo chất còn lại

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

\(m_{HCl}=\dfrac{150.18,25}{100}=27,375\left(g\right)\)

\(n_{HCl}=\dfrac{27,375}{36,5}=0,75\left(mol\right)\)

PTHH :

                \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

trc p/u :   0,3       0,75        

p/u:          0,3     0,6           0,3        0,3 

sau p/u :  0         0,15         0,3       0,3 

---> Sau p/ư HCl dư 

\(a,V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(b,m_{ddHCl}=0,6.36,5=21,9\left(g\right)\)

\(c,m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

\(m_{ddZnCl_2}=19,5+150-\left(0,3.2\right)=168,9\left(g\right)\)

\(C\%=\dfrac{40,8}{168,9}.100\%\approx24,16\%\)