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

nP=\(\dfrac{62}{31}\)=0,2(mol)

nO2=\(\dfrac{7,84}{22,4}\)=0,35(mol)

PTHH:4P+5O2to→2P2O5

tpứ:    0,2     0,35     

pứ:     0,2     0,25     0,1

spứ:     0     0,1         0,1

a)chất còn dư là oxi

mO2dư=0,1.32=3,2(g)

b)mP2O5=n.M=0,1.142=14,2(g)

\(a.n_P=0,2\left(mol\right);n_{O_2}=0,35\left(mol\right)\\ 4P+5O_2-^{t^o}\rightarrow2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,35}{5}\\ \Rightarrow SauphảnứngO_2dư\\ n_{O_2\left(pứ\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\\ \Rightarrow m_{P\left(dư\right)}=\left(0,35-0,25\right).32=3,2\left(g\right)\\ b.n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\\ \Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

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) 

PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{3}< \dfrac{0,4}{2}\) \(\Rightarrow\) Oxi còn dư, Fe p/ứ hết

\(\Rightarrow n_{O_2\left(dư\right)}=0,4-\dfrac{2}{15}=\dfrac{4}{15}\left(mol\right)\)

+) Theo PTHH: \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{2}{15}\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{15}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{kk}=\dfrac{2}{15}\cdot22,4\cdot5\approx14,93\left(l\right)\\m_{Fe_3O_4}=\dfrac{1}{15}\cdot232\approx15,47\left(g\right)\end{matrix}\right.\) 

4Na + O₂ -> 2Na₂O

 n_Na = == 0,4 ( mol )

( mol )

Lập tỉ lệ :

Ta có : 

Nên : O₂ dư

phản ứng = n_{Na * =} 0,4 / 4 = 0,1 ( mol )

dư = 0,2 - 0,1 = 0,1 ( mol )

= n * M = 0,1 * 32 = 3,2 ( g)

a)

\(n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)\\ n_{O_2} = \dfrac{3,36}{22,4}= 0,15(mol)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\)

Ta thấy : 

\( \dfrac{n_{Mg}}{2} = 0,1 < n_{O_2} = 0,15 \) nên O₂ dư.

\(n_{O_2\ pư} = \dfrac{n_{Mg}}{2} = 0,1(mol)\\ m_{O_2\ dư} = (0,15-0,1).32 = 1,6(gam)\\ V_{O_2\ dư} = (0,15-0,1).22,4 = 1,12(lít)\)

b)

\(n_{MgO} = n_{Mg} = 0,2\ mol\\ \Rightarrow m_{MgO} = 0,2.40 = 8\ gam\)

\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2mol\)

4P + 5O₂ \(\underrightarrow{t^o}\) 2P₂O₅

\(\dfrac{0,1}{4}< \dfrac{0,2}{5}\) => O₂ dư, Photpho đủ

\(n_{O_2}=0,2-0,04=0,16\left(mol\right)\)

\(m_{P_2O_5}=\) 0,05 . 142 = 7,1 ( g )