PTHH:

2KMnO₄ -to> K₂MnO₄ + MnO₂ + O₂

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

n KMnO4=\(\dfrac{31,6}{158}\)=0,2 mol

=>VO2=0,1.22,4=2,24l

b)2Mg+O₂-to>2MgO

     0,2----0,1----0,2

n Mg=\(\dfrac{7,2}{24}\)=0,3 mol

=>Mg dư :0,1 mol

=>mMg=0,1.24=2,4g

=>m MgO=0,2.40=8g

nKMnO4 = 15.8/158 = 0.1 (mol) 

2KMnO4 -to-> K2MnO4 + MnO2 + O2 

0.1__________________________0.05 

nNa = 1.38/23 = 0.06 (mol) 

4Na +  O2 -to-> 2Na2O 

0.06__0.015 

mO2 (dư) = ( 0.025 - 0.015) * 32 = 0.32(g)

n KClO3 = 4,9/122,5 = 0,04(mol)

$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$

n O2 = 3/2 n KClO3 = 0,06(mol)

X cho vào HCl thấy thoát ra khí chứng tỏ X chứa R dư

Gọi n là hóa trị của R

n H2 = 1,344/22,4 = 0,06(mol)

$4R + nO_2 \xrightarrow{t^o} 2R_2O_n$

$2R + 2nHCl \to 2RCl_n + nH_2$

n R = 4/n n O2 + 2/n n H2 = 0,36/n(mol)

Bảo toàn khối lượng :

=> m R = m X - m O2 = 6,24 - 0,06.32 = 4,32(gam)

Suy ra : 

0,36/n  . R = 4,32

=> R = 12n 
Với n = 2 thì R = 24(Magie)

a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Ta có: \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,3.24,79=7,437\left(g\right)\)

b, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)

Ta có: \(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,3}{1}\), ta được O₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,25\left(mol\right)\\n_{CuO}=n_{Cu}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

\(m_{CuO}=0,5.80=40\left(g\right)\)

a) \(n_{KMnO_4}=\dfrac{47,4}{158}=0,3\left(mol\right)\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄+ MnO₂ + O₂

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

=> V_O2 = 0,15.22,4 = 3,36 (l)

b) 

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

Xét tỉ lệ: \(\dfrac{0,1}{4}< \dfrac{0,15}{5}\) => P hết, O₂ dư

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

             0,1------------>0,05

            P₂O₅ + 3H₂O --> 2H₃PO₄

             0,05-------------->0,1

=> \(C_{M\left(H_3PO_4\right)}=\dfrac{0,1}{0,08}=1,25M\)

a, \(n_{KMnO_4}=\dfrac{47,4}{158}=0,3\left(mol\right)\)

PTHH: 2KMnO₄ ----t^o---> K₂MnO₄ + MnO₂ + O₂

Mol:        0,3                                                    0,15

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

b, \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

Ta có: \(\dfrac{0,1}{4}< \dfrac{0,15}{5}\) ⇒ P hết, O₂ dư

PTHH: 4P + 5O₂ ----t^o---> 2P₂O₅

Mol:    0,1                             0,05

PTHH: P₂O₅ + 3H₂O ----> 2H₃PO₄

Mol:     0,05                            0,1

\(C_{M_{H_3PO_4}}=\dfrac{0,1}{0,08}=1,25M\)

a)\(n_{O_2}=\dfrac{10,08}{22,4}=0,45mol\)

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

0,9                                                 0,45

\(m_{KMnO_4}=0,9\cdot158=142,2g\)

b)\(m_{K_2MnO_4}=0,45\cdot197=88,65g\)

c)\(2Fe+O_2\underrightarrow{t^o}2FeO\)

   0,9      0,45

\(m_{Fe}=0,9\cdot56=50,4g\)

= 50,4 g

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

\(nO_2=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

\(nKMnO_4=0,45.2=0,9\left(mol\right)\)

\(mKMnO_4=0,9.\left(39+55+16.4\right)=142,2\left(g\right)\)

\(nK_2MnO_4=nO_2=0,45\left(mol\right)\)

\(mK_2MnO_4=0,45.\left(39.2+55+16.4\right)=88,65\left(g\right)\)

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

3mol   2mol    1mol 

\(nO_2=0,45\left(mol\right)\)

=> \(nFe=0,45.\dfrac{3}{2}=0,675\left(mol\right)\)

=> \(mFe=0,675.56=37,8\left(g\right)\)

\(n_{KMnO_4}=\dfrac{15.8}{158}=0.1\left(mol\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(n_{O_2}=\dfrac{0.1}{2}=0.05\left(mol\right)\)

\(V_{O_2}=0.05\cdot22.4=1.12\left(l\right)\)

\(b.\)

\(n_P=\dfrac{6.2}{31}=0.2\left(mol\right)\)

\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)

\(4.........5\)

\(0.2........0.05\)

\(LTL:\dfrac{0.2}{4}>\dfrac{0.05}{5}\Rightarrow Pdư\)

\(m_{P\left(dư\right)}=\left(0.2-0.04\right)\cdot31=4.96\left(g\right)\)

PTHH:2KMnO4--- K2MnO4+MnO2 +O2

ADCT nKmno4=15,8/158=0,1 mol 

a, theo pt có nO2/nKmno4= 1/2 

nO2=0,05 mol 

ADCT V=n*22,4 

VO2=0,05*22,4 =1,12 l

b, PTHH: 5O2+4P---2P2O5

ADCTnP=6,2/31=0,2 mol 

Theo pt 

nO2/5=0,01     bé hơn     nP/4=0,05 

P dư 

theo pt nP(pư)/nO2=4/5 

nP(p/ư)=0,04 mol 

nP(dư)=0,05-0,04 =0,01 mol

ADCT:m=n*M

mP(dư)=0,01*31=0,31g

\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)

\(0.1.....0.05\)

\(m_{O_2\left(dư\right)}=\left(0.5-0.05\right)\cdot32=14.4\left(g\right)\)