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3H2+Fe2O3----->3H2O+2Fe
Ta có
n Fe=11,2/56=0,2(mol)
Theo pthh
n Fe2O3=1/2n Fe=0,1(mol)
m Fe2O3=0,1.160=16(g)
n H2=3/2n Fe=0,15(mol)
V H2=0,15.22,4=3,36(l)
a) PTHH: Fe2O3 + 3H2 =(nhiệt)=> 2Fe + 3H2O
nFe = \(\frac{42}{56}=0,75\left(mol\right)\)
=> nFe2O3 = \(\frac{0,75}{2}=0,375\left(mol\right)\)
=> mFe2O3(phản ứng) = 0,375 x 160 = 60 (gam)
b) Theo phương trình, nH2O = \(\frac{0,75\times3}{2}=1,125\left(mol\right)\)
=> nH2O(tạo thành) = 1,125 x 18 = 20,25 (gam)
a)Fe2O3+3H2=>3H2O+2Fe
nFe=42/56=0,75 mol
Từ pthh=>nFe2O3=0,375 mol=>mFe2O3=0,375.160=60gam
b)nH2O=1,125 mol=>mH2O=1,125.18=20,25gam
a) Fe2O3+3H2--->2Fe+3H2O
n Fe=79/56=1,4(mol)
Theo pthh
n Fe2O3=1/2n Fe=0,7(mol)
m Fe2O3=0,7.160=112(g)
b) n H2O=3/2n Fe=0,933(mol)
m H2O=0,933.18=16,794(g)
c) n H2=3/2n Fe=0,933(mol)
V H2=0,933.22,4=20,8992(l)
a)
\(n_{Fe}=\frac{79}{56}\left(mol\right)\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
79/112_237/112 __79/56__237/112
\(m_{Fe2O3}=\frac{160.79}{112}=112,86\left(g\right)\)
b)
\(m_{H2O}=\frac{237}{112.18}=38,09\left(g\right)\)
c)
\(\rightarrow V_{H2}=\frac{237}{112}.22,4=47,4\left(l\right)\)
a) PTHH: 4Al + 3O2 =(nhiệt)=> 2Al2O3
nAl = \(\frac{5,4}{27}=0,2\left(mol\right)\)
b) nO2 = \(\frac{0,2\times3}{4}=0,15\left(mol\right)\)
=> VO2(đktc) = 0,15 x 22,4 = 3,36 lít
c) nAl2O3 = \(\frac{0,2\times2}{4}=0,1\left(mol\right)\)
=> mAl2O3 = 0,1 x 102 = 10,2 gam
nFe=5,6/56=0,1(mol)
pt: Fe2O3+3H2--->2Fe+3H2O
0,05____________0,1
mFe2O3=0,05.160=8(g)
b) nO2=2,24/22,4=0,1(mol)
3Fe+2O2--->Fe3O4
3____2
0,1___0,1
Ta có: 0,1/3<0,1/2
=>O2 dư
Theo pt: nFe3O4=1/3nFe=1/3.0,1=0,033(mol)
=>mFe3O4=0,033.232=7,656(g)
Câu 2: nFe3O4=69,6/232=0,3(mol)
pt: Fe3O4+4H2--->3Fe+4H2O
0,3________1,2____0,9
VH2=1,2.22,4=26,88(l)
mH2=1,2.2=2,4(g)
mFe=0,9.56=50,4(g)
Bài 1
\(Fe2O3+3H2-->2Fe+3H2O\)
\(HgO+H2-->Hg+O2\)
\(PbO+H2-->Pb+H2O\)
Bài 2
a)\(Fe2O3+3H2-->2Fe+3H2O\)
\(CuO+H2-->Cu+H2O\)
\(m_{Fe2O3}=20.60\%=12\left(g\right)\)
\(n_{Fe2O3}=\frac{12}{160}=0,075\left(mol\right)\)
\(n_{Fe}=2n_{Fe2O3}=0,15\left(mol\right)\)
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
\(m_{CuO}=20-12=8\left(g\right)\)
\(n_{CuO}=\frac{8}{80}=0,1\left(mol\right)\)
\(n_{Cu}=n_{CuO}=0,1\left(mol\right)\)
\(m_{Cu}=0,1.64=6,4\left(g\right)\)
b)\(n_{H2\left(1\right)}=3n_{Fe2O3}=0,225\left(mol\right)\)
\(n_{H2\left(2\right)}=n_{CuO}=0,1\left(mol\right)\)
\(\sum n_{H2}=0,1+0,225=0,325\left(mol\right)\)
\(V_{H2}=0,325.22,4=7,28\left(l\right)\)
Bài 3
\(2H2+O2-->2H2O\)
\(n_{H2}=\frac{8,4}{22,4}=0,375\left(mol\right)\)
\(n_{O2}=\frac{2,8}{22,4}=0,125\left(mol\right)\)
Lập tỉ lệ
\(n_{H2}\left(\frac{0,375}{2}\right)>n_{O2}\left(\frac{0,125}{1}\right)=>H2dư\)
\(n_{H2O}=2n_{O2}=0,225\left(mol\right)\)
\(m_{H2O}=0,25.18=4,5\left(g\right)\)
Bài 1 :
nFe = 22.4/56=0.4 mol
Fe3O4 + 4H2 -to-> 3Fe + 4H2O
2/15_____8/15______0.4____8/15
VH2 = 8/15*22.4= 11.95 (l)
mH2O = 8/15*18=9.6 g
C1:
mFe3O4 = 2/15*232=30.93 g
C2:
Áp dụng ĐLBTKL :
mFe3O4 + mH2 = mFe + mH2O
m + 16/15 = 22.4 + 9.6
=> m = 30.93 g
Bài 2 :
nMg = 12/24=0.5 mol
nCu = 16/64=0.25 mol
Mg + 1/2O2 -to-> MgO
0.5____0.25_______0.5
Cu + 1/2O2 -to-> CuO
0.25___0.125_____0.25
VO2 = ( 0.25 + 0.125) *22.4 = 8.4 (l)
mMgO = 0.5*40=20 g
mCuO = 0.25*80=20 g
$a.PTHH :$
$2Fe(OH)_3\overset{t^O}\to Fe_2O_3+3H_2O$
$b.n_{Fe(OH)_3}=\dfrac{32,1}{107}=0,3mol$
$Theo$ $pt :$
$n_{Fe_2O_3}=\dfrac{1}{2}.n_{Fe_2O_3}=\dfrac{1}{2}.0,3=0,15mol$
\(\Rightarrow\)$m_{Fe_2O_3}=0,15.160=24g$
Fe2O3+3H2-to->2Fe+3H2O
0,05-------0,15-----------0,1 mol
nFe=5,6\56=0,1 mol
=>mFe2O3=0,05.160=8g
=>VH2=0,15.22,4=3,36l
Em mơn chị Linh nhìu lém ạ. :))