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

...9,2 gam chất gì :) ?

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)

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

a) n\(_P\) = \(\dfrac{6,2}{31}\) = 0,2(mol); n\(_{O_2}\) = \(\dfrac{7,84}{22,4}\) = 0,35(mol).

Theo phương trình hóa học:

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

4        5          2 (mol)

0,2 \(\rightarrow\)0,25 \(\rightarrow\) 0,1(mol)

Tỉ lệ số mol: \(\dfrac{0,2}{4}< \dfrac{0,35}{5}\) \(\Rightarrow\) \(O_2\) phản ứng dư, P phản ứng hết.

\(n_{O_2}\)dư = 0,35 - \(\dfrac{0,2.5}{4}\) = 0,1(mol) \(\Rightarrow\) \(m_{O_2}\)dư = 0,1.32 = 3,2g

b) \(n_{P_2O_5}\) = 0,1.142 = 14,2g

4Na + O₂ -> 2Na₂O

 n_Na = 9,2/23= 0,4 ( mol )

nNa/4=0,4/4=0,1

nO2/1>nNa/4(0,2>0,1)

Nên : O₂ dư

nO2phản ứng = n_{Na *} 

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)

Bài 1:

\(a,2Cu+O_2\underrightarrow{t^o}2CuO\)

b, \(n_{O_2}=\dfrac{1,12}{32}=0,035mol\)

\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)

\(\dfrac{0,1}{2}>\dfrac{0,035}{1}\) => Cu dư, O₂ đủ

\(n_{Cu}\left(dư\right)=0,1-0,07=0,039\left(mol\right)\)

c, \(m_{CuO}=0,07.80=5,6g\)

Bài 2:

\(n_{Al}=\dfrac{13,5}{27}=0,5mol\)

\(n_{O_2}=\dfrac{6,67}{32}=0,21\left(mol\right)\)

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

\(\dfrac{0,5}{4}>\dfrac{0,21}{3}\) => Al dư, O₂ đủ

\(n_{Al_2O_3}=\dfrac{2}{3}.0,21=0,14\left(mol\right)\)

\(m_{Al_2O_3}=0,14.102=14,28g\)

Chưa làm câu ni à

khó nhở anh cũng không biết làm

khó qá nhở ?

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

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