\(n_{O_2}=\dfrac{V_{O_2}}{24,79}=\dfrac{6,21}{24,79}=0,25mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

0,2     0,25              0,1      ( mol )

\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,1.142=14,2g\)

\(m_P=n_P.M_P=0,2.31=6,2g\)

a) \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)

           0,4-->0,5----->0,2

b) \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)

c) \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

Bài 1:

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(4P+5O_2\rightarrow2P_2O_5\)

0,24.... 0,3 .... 0,12 (mol)

\(m_P=0,24.31=7,44\left(g\right)\)

\(m_{P_2O_5}=0,12.142=17,04\left(g\right)\)

Bài 2:

\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

\(4Al+3O_2\rightarrow2Al_2O_3\)

0,8 .... 0,6 ...... 0,4 (mol) 

\(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)

\(V_{O_2}=0,6.22,4=13,44\left(l\right)\)

nCu = 6,4/64 = 0,1 (mol)

PTHH: 2Cu + O2 -> (t°) 2CuO

Mol: 0,1 ---> 0,05 ---> 0,1

mCuO = 0,1 . 80 = 8 (g)

Vkk = 0,05 . 5 . 22,4 = 5,6 (l)

a) 4P + 5O₂ --t^o--> 2P₂O₅

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

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

           0,2-->0,25------->0,1

=> m_P2O5 = 0,1.142 = 14,2(g)

c) V_O2 = 0,25.22,4 = 5,6(l)

6,7 l o2 số lẻ vậy 

a) nP=0,1(mol)

PTHH: 4P + 5 O2 -to-> 2 P2O5

nO2= 5/4 . nP=5/4. 0,1= 0,125(mol)

=> V(O2,dktc)=0,125.22,4=2,8(l)

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

=>mP2O5=142.0,05=7,1(g)

\(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ a,n_{O_2}=3.0,5=1,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\\ V_{kk\left(đktc\right)}=33,6.5=168\left(l\right)\\ b,n_{CO_2}=n_{H_2O}=2.0,5=1\left(mol\right)\\ m_{CO_2}=44.1=44\left(g\right);m_{H_2O}=18.1=18\left(g\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{Ca\left(OH\right)_2}=n_{CO_2}=1\left(mol\right)\\ m_{Ca\left(OH\right)_2}=1.74=74\left(g\right)\\ m_{ddCa\left(OH\right)_2}=\dfrac{74.100}{10}=740\left(g\right)\)