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\(Fe_2O_3\left(0,15\right)+3H_2\rightarrow2Fe\left(0,3\right)+3H_2O\)
\(3Fe\left(0,3\right)+2O_2\rightarrow Fe_3O_4\left(0,1\right)\)
\(n_{Fe_3O_4}=\frac{23,2}{232}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,15.160=24\left(g\right)\)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
Ta có: \(n_{Fe_3O_4}=\frac{23,2}{232}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 -to-> 2Fe + 3H2O (1)
3Fe + 2O2 -to-> Fe3O4 (2)
Theo các PTHH và đề bài, ta có:
\(n_{Fe\left(2\right)}=3.n_{Fe_3O_4\left(2\right)}=3.0,1=0,3\left(mol\right)\)
\(n_{Fe\left(1\right)}=n_{Fe\left(2\right)}=0,3\left(mol\right)\)
\(n_{Fe_2O_3\left(1\right)}=\frac{n_{Fe\left(1\right)}}{2}=\frac{0,3}{2}=0,15\left(mol\right)\)
Ta có: \(a=m_{Fe_2O_3\left(1\right)}=0,15.160=24\left(g\right)\)
\(b=m_{Fe\left(1\right)}=0,3.56=16,8\left(g\right)\)
nFe = 16,8 : 56 = 0,3 (mol)
pthh :3 Fe + 2O2 -t--> Fe3O4
0,3--------------> 0,1 (mol)
=> mFe3O4 =0,1 . 232 = 23,2(G)
nH2 = 44,8 : 22,4 = 2 (g)
pthh : Fe3O4 + H2 -t--> Fe + H2O
LTL : 0,1 / 1 < 2 /1
=> H2 du
nH2 (pu) = nFe3O4 = 0,1 (mol)
=> nH2 (d) = 2-0,1 = 1,9 (mol)
mH2 (d) = 1,9 . 2 = 3,8 (g)
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)
Câu 13:
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:R_2O_3+3H_2\underrightarrow{t^o}2R+3H_2O\\ Theo.pt:n_{R_2O_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\\ M_{R_2O_3}=\dfrac{16}{0,1}=160\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow2R+16.3=160\\ \Leftrightarrow R=56\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow R.là.Fe\\ CTHH:Fe_2O_3\)
Bài 14:
\(n_{H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ PTHH:Fe+H_2SO_{4\left(loãng\right)}\rightarrow FeSO_4+H_2\uparrow\left(1\right)\\ Theo.pt\left(1\right):n_{Fe}=n_{H_2}=0,125\left(mol\right)\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\left(2\right)\\ Theo.pt\left(2\right):n_{Fe_2O_3}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}.0,125=\dfrac{1}{24}\left(mol\right)\\ m=m_{Fe_2O_3}=\dfrac{1}{24}.160=\dfrac{20}{3}\left(g\right)\\ n=n_{Fe}=0,125.56=7\left(g\right)\)
MX= 19.2=38
=> nN2=nNO2=x
mFe + mO = moxit
=> mO = moxit- mFe = 7,36-5,6=1,76 gam
=> nO = 0,11 mol
Fe0 →Fe3+ + 3e
0,1 0,3
O0 +2e →O2- N+5 + 3e → N+2 N+5 +1e → N+4
0,11 0,22 3x x x x
Bảo toàn e: 0,3 = 0,22+3x+x =0,22+4x =>x= 0,02
=> nkhí = 2x=0,04 mol
=> Vkhí = 0,04.22,4=0,896 lít=896 ml
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
$m_{Al} = 0,04.27 =1,08\ gam$
$m_{Fe} = 0,06.56 = 3,36\ gam$
a) \(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
\(PTHH:ZnO+H_2\underrightarrow{t^o}Zn+H_2O\\ \left(mol\right)...0,2\rightarrow...0,2......0,2..0,2\)
b) \(m_{Zn}=0,2.65=13\left(g\right)\)
c) \(n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\)
\(PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\)
Ban đầu: 0,2.......0,3
Phản ứng: 0,2.....0,1......0,2
Dư:......................0,2
Lập tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\left(0,1< 0,3\right)\)
\(m_{ZnO}=0,2.81=16,2\left(g\right)\)
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
\(n_{Al_2O_3}=\dfrac{2}{3}.0,21=0,14\left(mol\right)\)
\(m_{Al_2O_3}=0,14.102=14,28g\)
n H2O=\(\dfrac{3,6}{18}\)=0,2 mol
2H2O-đp->2H2+O2
0,2-------------------0,1 mol
2O2+3Fe-to>Fe3O4
0,1----------------0,05 mol
=>x=VO2=0,1.22,4=2,24l
=>y=m Fe3O4=0,05.232=11,6g
PTHH:
Fe2O3 + 3H2 --> (nhiệt độ) 2Fe + 3H2O (1)
3Fe + 2O2 --> ( nhiệt độ) Fe3O4 (2)
nFe3O4=23.2 : 232 = 0.1 (mol)
PTHH (2) => nFe= 3nFe3O4 = 0.1 * 3 = 0.3 (mol)
=> b = mFe= 0.3*56 = 16.8 (g)
PTHH (1) => nFe= 2nFe3O4= 0.3 : 2 = 0.15 (mol)
=> a = mFe2O3 = 0.15 * 160 = 24 (g)
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