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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)
\(\rightarrow V_{H2}=\frac{237}{112}.22,4=47,4\left(l\right)\)
1. Zn(OH)2 ----->ZnO + H2O
2. 2Ba + O2 ------>2BaO
3. CuO + O2 ------>ko xảy ra
4.2 Mn + O2 ------>2MnO
5. 4P + 5O2 ------>2P2O5
6. 2C2H6 + 7O2 ----->4CO2 + 6H2O
7. 2KMnO4 ------>K2MnO4 + MnO2 + O2
8. Ca(HCO3)2 ------>CaCO3 + CO2 + H2O
9. Ag + O2 ------>ko xảy ra
10. SO3 + O3 ------>ko xảy ra
11. P2O5 + O2 ------>ko xảy ra
12. 2KNO3 ------>2KNO2 + O2
1. Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
2. nFe2O3= 5/160=1/32 mol
nH2SO4= 0.075 mol
Lập tỉ lệ: 1/32 > 0.075/3 => Fe2O3 dư
nFe2O3 dư= 1/32 - 0.075/3= 1/160 mol
mFe2O3 dư= 1/160*160=1 g
3. nFe2(SO4)3= 0.075/3=1/40 mol
mFe2(SO4)3= 1/40*400=10g
Phương trình hóa học:
Fe2O3 + 3H2SO4 => Fe2(SO4)3 + 3H2O
nFe2O3 = m/M = 5/160 =1/32 (mol);
nH2SO4= 0.075 (mol)
Lập tỉ số: 1/32 > 0.075/3 => Fe2O3 dư, H2SO4 hết
nFe2O3 dư = 1/32 - 0.075/3= 1/160 (mol) mFe2O3 dư = n.M = 1/160x160 = 1
nFe2(SO4)3 = 0.075/3 =1/40 (mol)
mFe2(SO4)3 = n.M = 10 (g)
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
\(1.\\ PTHH:Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\\ n_{Fe}=\frac{16,8}{56}=0,3\left(mol\right)\\ m_{Fe_2O_3}=0,15.160=24\left(g\right)\\ m_{CO}=0,45.28=12,6\left(g\right)\\ V_{CO_2}=0,45.22,4=10,08\left(l\right)\)
\(2.\\ PTHH:Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ n_{H_2SO_4}=1,5\left(mol\right)\Rightarrow\left\{{}\begin{matrix}n_{Al_2O_3}=n_{Al_2\left(SO_4\right)_3}=0,5\left(mol\right)\\n_{H_2O}=1,5\left(mol\right)\end{matrix}\right.\\ m_{Al_2O_3}=0,5.102=51\left(g\right)\\ m_{H_2O}=18.1,5=27\left(g\right)\\ C_1:m_{Al_2\left(SO_4\right)_3}=0,5.342=171\left(g\right)\\ C_2:m_{Al_2\left(SO_4\right)_3}=51+1,5.98-27=171\left(g\right)\)
a,
PTHH
\(Fe_2O_3+3H_2-->2Fe+3H_2O\)
b,
Áp dụng ĐLBTKL :
\(m_{Fe_2O_3}+m_{H_2}=m_{Fe}+m_{H_2O}\)
\(=>m_{Fe_2O_3}=m_{Fe}+m_{H_2O}-m_{H_2}=21+9-3=27\left(g\right)\)
Vậy ...
$a.PTHH :$
$2Fe(OH)_3\overset{t^O}\to Fe_2O_3+3H_2O$
$b.n_{Fe(OH)_3}=\dfrac{32,1}{107}=0,3mol$
$Theo$ $pt :$
$n_{Fe_2O_3}=\dfrac{1}{2}.n_{Fe_2O_3}=\dfrac{1}{2}.0,3=0,15mol$
\(\Rightarrow\)$m_{Fe_2O_3}=0,15.160=24g$
Đặt nFe2O3=a
nCuO=b
Ta có:
\(\left\{{}\begin{matrix}160a+80b=32\\112a+64b=24\end{matrix}\right.\)
=>a=0,1;0,2
mFe2O3=160.0,1=16(g)
mCuO=32-16=16(g)
nO=0,1.3+0,2=0,5(mol)
Ta có:
nO=nH2=0,5(mol)
VH2=22,4.0,5=11,2(lít)
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