\(n_{Al} = a ; n_{Fe} = b\Rightarrow 27a + 56b = 27,6(1)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,5a(mol)\\ n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = \dfrac{b}{3}(mol)\\ \Rightarrow 0,5a.102 + \dfrac{b}{3}232 = 43,6(2)\\ (1)(2) \Rightarrow a = 0,4 ; b = 0,3\\ m_{Al} = 0,4.27 = 10,8(gam) ; m_{Fe} = 0,3.56 = 16,8(gam)\)

Đặt \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

Theo đề: \(m_{hh}=39\left(g\right)\)

\(\Rightarrow m_{Al}+m_{Fe}=39\\ \Rightarrow27x+56y=39\left(1\right)\)

\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ \left(mol\right)....x\rightarrow..0.75x....0,5x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow..\dfrac{2}{3}y.....\dfrac{1}{3}y\)

Theo đề: \(n_{O_2}=\dfrac{V}{22,4}=\dfrac{12,32}{22,4}=0,55\left(mol\right)\)

\(\Rightarrow0,75x+\dfrac{2}{3}y=0,55\left(2\right)\)

\(\xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}27x+56y=39\\0,75x+\dfrac{2}{3}y=0,55\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,6.56=33,6\left(g\right)\end{matrix}\right.\\ m_r=m_{Al_2O_3}+m_{Fe_3O_4}=0,5.0,2.102+\dfrac{1}{3}.0,6.232=56,6\left(g\right)\)

\(m_{Al}=27,8.19,2\%=5,4\left(g\right)\\ m_{Fe}=27,8-5,4=22,4\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\end{matrix}\right.\)

PTHH:

4Al + 3O₂ --t^o--> 2Al₂O₃

0,2-->0,15------->0,1

3Fe + 2O₂ --t^o--> Fe₃O₄

0,4-->4/15--------->2/15

\(\rightarrow\left\{{}\begin{matrix}V_{kk}=\left(0,15+\dfrac{4}{15}\right).22,4.5=\dfrac{140}{3}\left(l\right)\\m_{Cran}=0,1.102+\dfrac{2}{15}.232=\dfrac{617}{15}\left(g\right)\end{matrix}\right.\)

mAl=27,8.19,42%=5,4g

⇒nAl=\(\dfrac{5,4}{27}\)=0,2mol

⇒nFe=\(\dfrac{27,8-5,4}{56}\)=0,4mol

4Al+3O2to→2Al2O34

3Fe+2O2to→Fe3O4

⇒nO2=\(\dfrac{3}{4}\)nAl+\(\dfrac{2}{3}\)nFe=\(\dfrac{5}{12}\)mol

⇒Vkk=\(\dfrac{5}{12}\).22,4.5=46,67l

b,

mrắn=27,8+mO2=27,8+​​\(\dfrac{5}{12}\)32=41,1g

→⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩Vkk=(0,15+415

a, Giả sử: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 27x + 56y = 13,8 (1)

BTNT Al và Fe, có: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}x\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{2}x.102+\dfrac{1}{3}y.232=21,8\left(g\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{13,8}.100\%\approx39,1\%\\\%m_{Fe}\approx60,9\%\end{matrix}\right.\)

b, BTNT O, có: \(n_{O_2}=\dfrac{3n_{Al_2O_3}+4n_{Fe_3O_4}}{2}=0,6\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)

Bạn tham khảo nhé!

\(m_{Al}=19,2\%.27,8=5,3376\left(g\right)\Rightarrow n_{Al}=0,2\left(mol\right)\)

\(m_{Fe}=27,8-5,3376=22,4624\left(g\right)\Rightarrow n_{Fe}=0,4\left(mol\right)\)

\(4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\)

\(3Fe+2O_2-^{t^o}\rightarrow Fe_3O_4\)

Theo PT : \(n_{O_2}=0,2.\dfrac{3}{4}+0,4.\dfrac{2}{3}=\dfrac{5}{12}\left(mol\right)\)

Vì oxi chiếm 20% thể tích không khí

=> \(V_{kk}=\dfrac{5}{12}.22,4.\dfrac{100}{20}=\dfrac{140}{3}\left(lít\right)=46,67\left(lít\right)\)

Bảo toàn khối lượng ta có: \(m_{KL}+m_{O_2}=m_{oxit}\)

=> \(m_{oxit}=27,8+\dfrac{5}{12}.32=\dfrac{617}{15}\left(g\right)=41,13\left(g\right)\)

a, 

mAl=27,8.19,42%=5,4gmAl=27,8.19,42%=5,4g

⇒nAl=5,427=0,2mol⇒nAl=5,427=0,2mol

⇒nFe=27,8−5,456=0,4mol⇒nFe=27,8−5,456=0,4mol

4Al+3O2to→2Al2O34Al+3O2→to2Al2O3

3Fe+2O2to→Fe3O43Fe+2O2→toFe3O4

⇒nO2=34nAl+23nFe=512mol⇒nO2=34nAl+23nFe=512mol

⇒Vkk=512.22,4.5=46,67l⇒Vkk=512.22,4.5=46,67l

b,

mrắn=27,8+mO2=27,8+512.32=41,1g

\(m_{Al}=27,8.\dfrac{19,2}{100}=5,4\) ( g )

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\)  ( mol ) 

\(m_{Fe}=27,8-5,4=22,4\) ( g )

\(n_{Fe}=\dfrac{m}{m}=\dfrac{22,4}{56}=0,4\) ( mol )

\(4Al\) + \(3O_2\) →  \(2Al_2O_3\)

\(3Fe\) + \(2O_2\) →  \(Fe_3O_4\)

\(n_{O_2}=\dfrac{3}{4}Al+\dfrac{2}{3}Fe=\dfrac{3}{4}.0,2+\dfrac{2}{3}.0,4=\dfrac{5}{12}\)  ( mol )

\(V_{kk}=5V_{O_2}=5.\dfrac{5}{12}.22,4\approx46,7\) ( \(l\) )

0,2 mol                    →       0,1 mol

a

PTHH:

\(2Al\left(OH\right)_3\underrightarrow{t^o}Al_2O_3+3H_2O\)

\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

b

Gọi \(\left\{{}\begin{matrix}n_{Al\left(OH\right)_3}=x\left(mol\right)\\n_{Fe\left(OH\right)_3}=y\left(mol\right)\end{matrix}\right.\)

Theo PTHH suy ra: \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,5x\left(mol\right)\\n_{Fe_2O_3}=0,5y\left(mol\right)\end{matrix}\right.\)

Theo đề có hệ phương trình: \(\left\{{}\begin{matrix}78x+107y=29,2\\102.0,5x+160.0,5y=21,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al\left(OH\right)_3}=\dfrac{78.0,1.100\%\%}{29,2}=26,71\%\\\%m_{Fe\left(OH\right)_3}=\dfrac{107.0,2.100\%}{29,2}=73,29\%\end{matrix}\right.\)

Gọi $n_{Al}= a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 4,44(1)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$
$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$

B gồm : $Al_2O_3, Fe$

$n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,5a(mol)$

Suy ra: $0,5a.102 + 56b = 5,4(2)$
Từ (1)(2) suy ra a = 0,04 ; b = 0,06

$m_{Al} = 0,04.27 =1,08\ gam$

$m_{Fe} = 0,06.56 = 3,36\ gam$

Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

Theo đề: \(m_{hh}=36\left(g\right)\)

\(\Rightarrow m_{Mg}+m_{Fe}=36\\ \Rightarrow24x+56y=36\left(1\right)\)

\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)....x\rightarrow...0,5x.....x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow...\dfrac{2}{3}y....\dfrac{1}{3}y\)

Theo đề: \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(\Rightarrow0,5x+\dfrac{2}{3}y=0,6\left(2\right)\)

\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}24x+56y=36\\0,5x+\dfrac{2}{3}y=0,6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,8.24=19,2\left(g\right)\\m_{Fe}=0,3.56=16,8\left(g\right)\end{matrix}\right.\\ m_r=m_{MgO}+m_{Fe_3O_4}=0,8.40+\dfrac{1}{3}.0,3.232=55,2\left(g\right)\)

PTHH: 

mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)

VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)

mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)