a.Mg + 1/2O2 -> MgO

b.\(nMg=\dfrac{4.8}{24}=0.2mol\) => \(nO2=0.2\times\dfrac{1}{2}=0.1mol\)

\(V_{O2}=0.1\times22.4=2.24l\)

c.\(nMgO=nMg=0.2mol\)

\(mMgO=0.2\times40=8g\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

0,1        0,05          0,1    ( mol )

\(V_{O_2}=0,05.22,4=1,12l\)

\(m_{MgO}=0,1.40=4g\)

nMg= 2,4 : 24 = 0,1 ( mol ) 
pthh 2 Mg + O2 -t--> 2MgO 
        0,1 ---> 0,05 --->0,1 (mol 
=> VO2 = 0,05 . 22,4 = 1,12 (l) 
=> mMgO = 0,1 . 40 = 4 (G) 

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(2Mg+O_2\underrightarrow{^{^{t^0}}}2MgO\)

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

\(V_{O_2}=0.1\cdot22.4=2.244\left(l\right)\)

\(m_{MgO}=0.2\cdot40=8\left(g\right)\)

\(a) 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ b) n_{O_2} = \dfrac{1}{2}n_{Mg} = \dfrac{1}{2}.\dfrac{12}{24} = 0,25(mol)\\ \Rightarrow V_{O_2} = 0,25.22,4 = 5,6(lít)\\ V_{không\ khí} = 5V_{O_2} = 5,6.5 = 28(lít)\\ c) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5(mol)\\ m_{KMnO_4} = 0,5.158 =79(gam)\)

a) $2Mg + O_2 \xrightarrow{t^o} 2MgO$

Theo PTHH : n MgO = n Mg = 2,4/24 = 0,1(mol)

=> m MgO = 0,1.40 = 4(gam)

b) n O2 = 1/2 n Mg = 0,05(mol) 

=> V O2 = 0,05.22,4 = 1,12(lít)

=> V không khí = 5V O2 = 1,12.5 = 5,6(lít)

c) $Mg + 2HCl \to MgCl_2 + H_2$

n HCl = 200.15%/36,5 = 60/73(mol)

Ta thấy :

n Mg / 1 = 0,1 < n HCl / 2 = 30/73 suy ra HCl dư

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

=> m dd sau pư = 2,4 + 200 - 0,1.2 = 202,2(gam)

Vậy :

C% MgCl2 = 0,1.95/202,2  .100% = 4,7%

C% HCl = (60/73 - 0,1.2).36,5/202,2  .100% = 11,23%

mMg = 3.6/24 = 0.15 (mol) 

2Mg + O2 -to-> 2MgO 

0.15__0.075____0.15 

mMgO= 0.15*40 = 6 (g) 

VO2  = 0.075*22.4 = 1.68 (l) 

2KClO3 -to-> 2KCl + 3O2 

0.05_______________0.075 

mKClO3 = 0.05*122.5 = 6.125 (g) 

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

a+b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{MgO}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{MgO}=0,15\cdot40=6\left(g\right)\end{matrix}\right.\)

c) PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)

Theo PTHH: \(n_{KClO_3}=0,05\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=0,05\cdot122,5=6,125\left(g\right)\)

a.b.c.\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}\dfrac{4,8}{24}=0,2mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

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

\(V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24l\)

\(m_{MgO}=n_{MgO}.M_{MgO}=0,2.40=8g\)

d. Sửa đề: tính khối lượng KMnO4

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

    0,2                                                      0,1    ( mol )

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,2.158=31,6g\)

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

PTHH: \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

          0,2--->0,15

b) \(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

c) PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

                  0,3<-----------------------------------0,15

\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)