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

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 ^^

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

 1)

a) photpho+oxi--->điphotpho pentaoxit

b)M photpho+M oxi---> M điphotpho pentaoxit

3.1 + M oxi--> 7.1

M oxi = 7.1 - 3.1 = 4g

mới làm xong bài 1 bấm xem thêm thấy bất ngờ lun nên thôi ko lm nx hihi

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

Theo ĐLBTKL, ta có:

\(m_P+m_{O_2}=m_{P_2O_5}\\ \Rightarrow m_{O_2}=7,1-3,1=4g\)

Ta có : \(m_P+m_o=m_{p_2O_5}\)

hay  \(3,1+5O_2=7,1\)

       \(\Rightarrow m_{O_2}=7,1=3,1=4\left(gam\right)\)

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

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

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

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

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

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

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

           0,2-->0,25------->0,1

=> \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

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

Bài 2:

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

PTHH: 4P + 5O₂ → 2P₂O₅

Mol:     0,4                    0,2

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

Bài 1:

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH: 4Al + 3O₂ ---t^o→ 2Al₂O₃

Mol:    0,4     0,3

\(V_{O_2}=0,3.22,4=6,72\left(l\right)\)

\(n_{O_2\left(đktc\right)}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ 0,12........0,15.........0,06\left(mol\right)\\ m_P=0,12.31=3,72\left(g\right)\)

Ta có: \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o---> 2P₂O₅.

Theo PT: n_P = \(\dfrac{4}{5}.n_{O_2}=\dfrac{4}{5}.0,15=0,12\left(mol\right)\)

=> m_P = 31 . 0,12 = 3,72(g)

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)

\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)