4P+5O2-to>2P2O5

0,4---0,5-----0,2

n P=0,4 mol

=>VO2=0,5.22,4=11,2l

=>Vkk=1,12.5=5,6l

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

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

\(\Rightarrow n_{O_2}=0,5\left(mol\right)\) \(\Rightarrow V_{O_2}0,5\cdot22,4=11,2\left(l\right)\)

a) \(4P+5O_2\underrightarrow{t\text{°}}P_2O_5\)

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

Từ PTHH: \(n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,4=0,5\left(mol\right)\)

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

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

a.b.\(n_{Mg}=\dfrac{m}{M}=\dfrac{6,4}{24}=\dfrac{4}{15}mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

4/15     2/15                       ( mol )

\(V_{O_2}=n.22,4=\dfrac{2}{15}.22,4=\dfrac{224}{75}l\)

c.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

       4/15                                                   2/15 ( mol )

\(m_{KMnO_4}=n.M=\dfrac{4}{15}.158=\dfrac{632}{15}g\)

nMg = 6,4 : 24= 0,26(mol) 
pthh : 2Mg+O2 -t--> 2MgO
       0,26 --> 0,13 (mol ) 
=> VO2(đktc) = 0,13.22,4=2,912(l)
pthh : 2KMnO4-t--> K2MnO4 + MnO2+ O2
                 0,26<------------------------------0,13(mol) 
 => mKMnO4 = 0,26.158= 41,08(g)

a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

b, \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)

c, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)

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

Phần b phải là thể tích O2 bạn nhé!

a)

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

Sản phẩm : Điphotpho pentaoxit.

b)

\(n_P = \dfrac{6,2}{31} = 0,2(mol)\\ \Rightarrow n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol)\\ \Rightarrow m_{P_2O_5} = 0,1.142 = 14,2(gam)\)

c)

\(n_{O_2} = \dfrac{5}{4}n_P = 0,125(mol)\\ \Rightarrow V_{O_2} = 0,125.22,4 = 2,8(lít)\)

d)

\(V_{không\ khí} = \dfrac{2,8}{20\%} = 14(lít)\)

Giúp mình với mình đang cần gấp😇

\(n_C=\dfrac{14,4}{44}=\dfrac{18}{55}\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ n_{O_2}=n_C=n_{CO_2}=\dfrac{18}{55}\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=\dfrac{18}{55}.22,4=\dfrac{2016}{275}\left(lít\right)\\ b,V_{kk}=\dfrac{100}{21}.\dfrac{2016}{275}=\dfrac{381}{11}\left(lít\right)\\ c,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3\left(LT\right)}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.\dfrac{18}{55}=\dfrac{12}{55}\left(mol\right)\\ \Rightarrow n_{KClO_3\left(TT\right)}=120\%.\dfrac{12}{55}=\dfrac{72}{275}\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.\dfrac{72}{275}=\dfrac{1764}{55}\left(g\right)\)