\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)

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

\(0.1.......0.125.....0.05\)

\(V_{O_2}=0.125\cdot22.4=2.8\left(l\right)\)

\(m_{P_2O_5}=0.05\cdot142=7.1\left(g\right)\)

n_P= 3,1 / 31 =0,1 mol

          2P +   5/2O₂ → P₂O₅

           0,1     0,125       0,05      mol

V_O2=0,125.22,4=2,8 l

b) mP₂O₅=0,05.142=7,1 g

a. \(n_P=\dfrac{3.1}{31}=0,1\left(mol\right)\)

PTHH : 4P + 5O₂ ---t^o----> 2P₂O₅

             0,1     0,125           0,05

b. \(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)

c. \(V_{O_2}=0,125.22,4=2,8\left(l\right)\\ \Rightarrow V_{kk}=2,8.5=14\left(l\right)\)

4P+5O2-to>2P2O5

0,1-----0,125----0,05

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

=>Vkk=0,125.22,4.5=14l

=>m P2O5=0,05.142=7,1g

TK

a. n P = 3.1 31 = 0 , 1 ( m o l ) PTHH : 4P + 5O2 ---to----> 2P2O5 0,1 0,125 0,05 b. m P 2 O 5 = 0 , 05.142 = 7 , 1 ( g ) c. V O 2 = 0 , 125.22 , 4 = 2 , 8 ( l ) ⇒ V k k = 2 , 8.5 = 14 ( l )

\(PTHH:4P+5O_2->2P_2O_5\)

Số mol của Photpho: \(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(PTHH:4P+5O_2->2P_2O_5\)

             4 mol   5 mol    2 mol

             0,2 mol --------> 0,1 mol

Khối lượng Diphotpho Pentaoxit: \(\left(M_{P_2O_5}=142g/mol\right)\)

\(m_{P_2O_5}=n.M=0,1.142=14,2\left(g\right)\)

b) \(PTHH:4P+5O_2->2P_2O_5\)

                 4 mol  5 mol

            0,2 mol -> 0,25 mol

Thể tích khí oxi (đktc) cần dùng: \(V_{O_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)

Chúc bn học tốt nha ^^

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\a, 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\b,n_{P_2O_5}=\dfrac{2}{5}.0,25=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=142.0,1=14,2\left(g\right)\\c,V_{kk\left(đktc\right)}=4.5,6=28\left(lít\right) \)

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

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

Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

c, \(n_{O_2}=\dfrac{5}{4}n_P=0,5\left(mol\right)\)

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

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

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)

nP = 6.2/31 = 0.2 (mol) 

4P + 5O2 -to-> 2P2O5

0.2......0.25

VO2 = 0.25*22.4 = 5.6 (l) 

Vkk = 5VO2 = 28 (l) 

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) 

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

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

\(a,PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5\\ b,n_P=\dfrac{6,2}{31}=0,2(mol)\\ \Rightarrow n_{O_2}=\dfrac{5}{4}n_P=0,25(mol)\\ \Rightarrow V_{O_2(đktc)}=0,25.22,4=5,6(l)\\ c,n_{P_2O_5}=\dfrac{1}{2}n_P=0,1(mol)\\ \Rightarrow m_{P_2O_5}=0,1.142=14,2(g)\)

 Ta có PTHH: 4P + 5O₂ -> 2P₂O₅

                        0,2---0,25 ----0,1 mol
nP = 6,2/31 = 0,2 mol:
b)

=>VO₂=0,25.22,4=5,6l
c)
=>mP₂O₅ = 0,1 . 142 = 14,2 (g)