\(a)\\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,05 < \dfrac{n_{O_2}}{5} = 0,06\)

Do đó, Oxi dư.

\(n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)\\ \Rightarrow m_{O_2\ dư} = (0,3 - 0,25).32 = 1,6(gam)\\ b)\\ n_{P_2O_5} = \dfrac{n_P}{2} = 0,1(mol)\\ \Rightarrow m_{P_2O_5} = 0,1.142 = 14,2(gam)\)

...9,2 gam chất gì :) ?

\(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ a.PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 

                    4         3            2

                   0,4      0,3         0,2

\(m_{Al_2O_3}=n.M=0,2.\left(27.2+16.3\right)=20,4\left(g\right)\\ c.V_{O_2}=n.24,79=0,3.24,79=7,437\left(l\right)\) 

\(d.n_{O_2}=\dfrac{m}{M}=\dfrac{12,8}{\left(16.2\right)}=0,4\left(mol\right)\\ PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 

                 4         3          2

              0,53      0,4      0,27

Tỉ lệ: \(\dfrac{0,53}{4}< \dfrac{0,4}{3}< \dfrac{0,27}{2}\Rightarrow Al_2O_3\) dư và dư \(m_{Al_2O_3}=n.M=0,27.\left(27.2+16.3\right)=27,54\left(g\right).\)

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

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

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

c, \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)

d, \(n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,4}{4}< \dfrac{0,4}{3}\), ta được O₂ dư.

\(\Rightarrow n_{O_2\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,1.32=3,2\left(g\right)\)

a) \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right);n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: 4Na + O₂ --t^o--> 2Na₂O

Xét tỉ lệ: \(\dfrac{0,4}{4}>\dfrac{0,05}{1}\) => Na dư, O₂ hết

PTHH: 4Na + O₂ --t^o--> 2Na₂O 

              0,2<-0,05------>0,1

=> \(m_{Na\left(dư\right)}=\left(0,4-0,2\right).23=4,6\left(g\right)\)

b) \(\left\{{}\begin{matrix}\%m_{Na\left(dư\right)}=\dfrac{4,6}{9,2+0,05.32}.100\%=42,6\%\\\%m_{Na_2O}=\dfrac{0,1.62}{9,2+0,05.32}.100\%=57,4\%\end{matrix}\right.\)

a: \(n_{Na}=\dfrac{9.2}{23}=0.4\left(mol\right)\)

\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

=>Na dư 0,35 mol

b: \(4Na+O_2\rightarrow2Na_2O\)

a: nNa=9.223=0.4(mol)nNa=9.223=0.4(mol)

nO2=1.1222.4=0.05(mol)nO2=1.1222.4=0.05(mol)

=>Na dư 0,35 mol

b: 4Na+O2→2Na2O4Na+O2→2Na2O

nFe = 2.8/56 = 0.05 (mol) 

nO2 = 22.4 / 22.4 = 1 (mol) 

3Fe + 2O2 -to-> Fe3O4 

0.05__1/30______1/60

mO2 (dư) = ( 1 - 1/30) * 32 = 30.93 (g) 

mFe3O4 = 1/60 * 232 = 3.867 (g) 

a/ \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b/ Ta có: \(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)

\(n_{O_2}=\dfrac{22.4}{22.4}=1\left(mol\right)\)

Ta có: \(\dfrac{n_{Fe\left(bra\right)}}{n_{Fe\left(pt\right)}}=\dfrac{0.05}{3}=0.016< \dfrac{n_{O_2\left(bra\right)}}{n_{O_2\left(pt\right)}}=\dfrac{1}{2}=0.5\)

=> Oxi phản ứng dư

mO2 dư = (1 - 1/30) . 32 = 30.93 (g) 

mFe3O4 = 1/60 . 232 = 3.867 (g) 

a) 3Fe + 2O_{2  t o}  Fe₃O₄

_b) n_Fe = \(\dfrac{2,8}{56}\)= 0,05 (mol)

n_O2= \(\dfrac{22,4}{22,4}\)= 1 (mol)

 3Fe +2 O₂ toFe₃O₄

\(\dfrac{3Fe}{2O_2}=\dfrac{0,05}{1}\Leftrightarrow\dfrac{Fe}{O_2}=\dfrac{0,1}{3}\)Vậy O₂ dư

           3Fe +2 O₂ toFe₃O₄

(mol)  0,05-> \(\dfrac{1}{30}\)-------> \(\dfrac{1}{60}\)

m_O2dư= (1-\(\dfrac{1}{30}\)₎. 32 = 30,93 (g)

c) m_Fe3O4=\(\dfrac{1}{60}\) . 232 = 3,87 (g)

4P + 5O₂ -----to---> 2P₂O₅

0,4---0,5-----------> 0,2 (mol)

+ n _P = 12,4 / 31 = 0,4 (mol)

+nO2 = 13,44 / 22,4 = 0,6 (mol)

Vì n_P/4 = 0,1 < n _O2 /5 = 0,12

=> Oxi còn thừa sau phản ứng .

m_{O2 dư} = (0,6 - 0,5 ) . 32 = 3,2 (g)

b. chất tạo thành : P₂O₅

m_P2O5 = 0,2 . ( 2.31 + 16 . 5 ) = 28,4 (g)

\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ pthh:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,4}{4}< \dfrac{0,6}{5}\) 
=> Oxi dư 
\(n_{O_2\left(p\text{ư}\right)}=\dfrac{5}{4}n_P=0,5\left(mol\right)\\ m_{O_2\left(d\right)}=\left(0,6-0,5\right).32=3,2\left(g\right)\\ n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\\ m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

nAl=\(\dfrac{5,4}{27}\)=0,2mol

nO2=\(\dfrac{4,48}{22,4}\)=0,2mol

PTHH:

4Al + 3O2--to->2Al2O3

Tỉ lệ \(\dfrac{0,2}{4}\) <\(\dfrac{0,2}{3}\)->Al hết O2 dưtính theo Al

=>m O2=\(\dfrac{1}{60}\).32=\(\dfrac{8}{15}\)g

2KMnO4-to>K2MnO4+MnO2+O2

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

m KMnO4=0,4.158=63,2g

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