2KMnO₄-to>K₂MnO₄+MnO₂+O₂

0,3-----------------0,15-----0,15------0,15 mol

n _KMnO4=\(\dfrac{47,4}{158}\)=0,3 mol

=>mcr=0,15.197.0,15.87=42,6g

=>V_O2=0,15.22,4=3,36l

b) 4P+5O₂-to>2P₂O₅

      0,1--------------0,05

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

->O2 dư

=>m _P2O5=0,05.142=7,1g

mKMnO4 = 47,4/158 = 0,3 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

Mol: 0,3 ---> 0,15 ---> 0,15 ---> 0,15

m = 0,15 . 197 + 0,15 . 87 = 85,2 (g)

V = VO2 = 0,15 . 22,4 = 3,36 (l)

nP = 3,1/31 = 0,1 (mol)

PTHH: 4P + 5O2 -> (t°) 2P2O5

LTL: 0,1/4 < 0,15/5 => O2 dư

nP2O5 = 0,1/2 = 0,05 (mol)

mP2O5 = 0,05 . 142 = 7,1 (g)

mKMnO4 = 47,4/158 = 0,3 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

Mol: 0,3 ---> 0,15 ---> 0,15 ---> 0,15

m = 0,15 . 197 + 0,15 . 87 = 42,6 (g)

V = VO2 = 0,15 . 22,4 = 3,36 (l)

nP = 3,1/31 = 0,1 (mol)

PTHH: 4P + 5O2 -> (t°) 2P2O5

LTL: 0,1/4 < 0,15/5 => O2 dư

nP2O5 = 0,1/2 = 0,05 (mol)

mP2O5 = 0,05 . 142 = 7,1 (g)

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

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

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

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

\(b.\)

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

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

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

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

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

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

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

\(a) n_P = \dfrac{12,4}{31} = 0,4(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_{P_2O_5} = \dfrac{1}{2}n_P = 0,2(mol)\\ \Rightarrow m_{P_2O_5} = 0,2.142 = 28,4(gam)\\ b) P_2O_5 + 3H_2O \to 2H_3PO_4\\ n_{H_3PO_4} = 2n_{P_2O_5} = 0,4(mol)\\ m_{dd} = 28,4 + 200 = 228,4(gam)\\ \Rightarrow C\%_{H_3PO_4} = \dfrac{0,4.98}{228,4}.100\% = 17,16\%\)

n_P = 12.4/31 = 0.4 (mol) 

4P + 5O₂ -t⁰-> 2P₂O₅

0.4.......................0.2

m_P2O5 = 0.2 * 142 = 28.4 (g) 

P₂O₅ + 3H₂O => 2H₃PO₄

0.2.............................0.4

m_H3PO4 = 0.4 * 98 = 39.2 (g) 

m_ddH3PO4 = 28.4 + 200 = 228.4 (g) 

C% H₃PO₄ = 39.2/228.4 * 100% = 17.16%

a)n_O2=\(\dfrac{3.36}{22.4}\)=0,15(mol)

2KMnO₄(t^o)→K₂MnO₄+MnO₂+O₂

Theo PT: n_KMnO4=2n_O2=0,3(mol)

→m=m_KMnO4=0,3.158=47,4(g)

b)n_H2=\(\dfrac{8.96}{22.4}\)=0,4(mol)

2H₂+O₂(t^o)→2H₂O

Vì \(\dfrac{nH_2}{2}\)<n_O2→O₂nH₂ dư

Theo PT: n_H2O=n_H2=0,4(mol)

→m_H2O=0,4.18=7,2(g)

Chọn đáp án D

\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2mol\)

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

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

a)\(V_{O_2}=0,1\cdot22,4=2,24l\)

b)\(m_{CRắn}=m_{K_2MnO_4}+m_{MnO_2}=0,1\cdot197+0,1\cdot87=28,4g\)

c)\(n_{CH_4}=\dfrac{11,2}{22,4}=0,5mol\)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

0,5        0,1    0              0

0,05      0,1    0,05         0,1

0,45       0      0,05         0,1

\(V_{CO_2}=0,05\cdot22,4=1,12l\)

\(m_{H_2O}=0,1\cdot18=1,8g\)

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

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

b, n_KMnO4= \(\dfrac{31,6}{158}\)=0,2 mol

Theo pt: n_O2=\(\dfrac{1}{2}\).0,2=0,1 mol

=> V_O2= 0,1.22,4= 2,24 l

=>m cr=0,1.197+0,1.87=28,4g

CH4+2O2-to>CO2+2H2O

            0,5-----0,25-----0,5

n CH4=\(\dfrac{11,2}{22,4}\)=0,5 mol

=>Oxi du

=>V CO2=0,25.22,4=5,6l

=>m H2O=0,5.18=9g

Đáp án : D

2KMnO₄ -> K₂MnO₄ + MnO₂ + O₂

S + O₂ -> SO₂

SO₂ + 2NaOH -> Na₂SO₃ + H₂O

SO₂ + NaOH -> NaHSO₃

Ta có : n_SO2 = n_O2 = ½ n_KMnO4  = 0,1 mol

Xét muối thu được gồm x mol Na₂SO₃ và y mol NaHSO₃

=> bảo toàn S : n_SO2 = x + y = 0,1

Và m_muối = 126x + 104y = 11,72g

=> x = 0,06 ; y = 0,04 mol

=> n_NaOH = 2n_Na2SO3 + n_NaHSO3 ( bảo toàn Na) = 0,16 mol

=> a = 1,6M

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

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

Ta có: \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{1}{2}\cdot\dfrac{31,6}{158}\cdot50\%=0,05\left(mol\right)\\n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,8}{4}>\dfrac{0,05}{5}\) \(\Rightarrow\) Photpho còn dư, Oxi p/ứ hết

\(\Rightarrow n_{P_2O_5}=0,02\left(mol\right)\) \(\Rightarrow m_{P_2O_5}=0,02\cdot142=2,84\left(g\right)\)

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

Ta có: \(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\)

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

Mà:  H% = 50%

\(\Rightarrow n_{O_2\left(TT\right)}=0,1.50\%=0,05\left(mol\right)\)

Ta có: \(n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,8}{4}>\dfrac{0,05}{5}\), ta được P dư.

Theo PT: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,02\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,02.142=2,84\left(g\right)\)

Bạn tham khảo nhé!

\(KMnO_4--t^o->K_2MnO_4+MnO_2+O_2\)

158                                                              32

31,6                                                             \(x\)

Với \(H=50\%\) \(\Rightarrow x=\dfrac{31,6.32}{158}.50\%=3,2\left(g\right)\)

\(4P+5O_2--t^o->2P_2O_5\)

124      160

\(y\)           3,2

\(\Rightarrow y=\dfrac{124.3,2}{160}=2,48\left(g\right)\)

\(n_{KMnO_4} = \dfrac{31,6}{158} = 0,2(mol) \Rightarrow n_{KMnO_4\ pư} = 0,2.50\% = 0,1(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{O_2} = \dfrac{n_{KMnO_4}}{2} = 0,05(mol)\\ n_P = \dfrac{24,8}{31} = 0,8(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,2 > \dfrac{n_{O_2}}{5} = 0,01 \to P\ dư\\ n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,02(mol)\\ m_{P_2O_5} = 0,02.142 = 2,84(gam)\)