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nFeO=48 : 72 = 0,67 (mol)
pthh : FeO + H2 -t-> Fe + H2O
0,67-->0,67--->0,67 (mol )
VH2 = 0,67 . 22,4 = 14,93 (l)
mFe = 0,67 . 56 = 37,52 (g)
Bài 3:
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
a, PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
0,1<------0,4
Zn + 2HCl ---> ZnCl2 + H2
0,4<-------------------------0,4
b, mFe3O4 = 0,1.232 = 23,2 (g)
c, mZn = 0,4.65 = 26 (g)
Bài 4:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a, PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1---->0,2---------------->0,1
b, VH2 = 0,1.22,4 = 2,24 (l)
c, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)
a.b.c.\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{9,8}{98}=0,1mol\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,1 0,1 0,1 ( mol )
\(V_{H_2}=n.22,4=0,1.22,4=2,24l\)
\(m_{Zn}=n.M=0,1.65=6,5g\)
d.\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{1,6}{160}=0,01mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,01 < 0,1 ( mol )
0,01 0,02 ( mol )
\(m_{Fe}=n.M=0,02.56=1,12g\)
a,PTHH:Zn+H2SO4-->ZnSO4+H2
b,nH2SO4=9,8÷(2×32+16×4)=0,1
PTHH:Zn+H2SO4--->ZnSO4+H2
0,1 0,1 0,1 0,1
-->Vh2=0,1×22,4=2,24
c,nZn=0,1
mZn=0,1×65=6,5
d,nH2=0,1,nfe2o3=1,6+(56×2+16×3)=0,01
PTHH:3H2+Fe2O3--->2Fe+3H2O
0,1 0,01
Ta thấy 0,1/3>0,1/1 nên sau pư h2 dư,fe2o3 hết.tính theo fe2o3
nfe=2nfe2o3=0,02
--->mfe=0,02×56=1,12
PT: Fe2O3+3H2to→2Fe+3H2O
CuO+H2to→Cu+H2O
a, Ta có: mFe2O3=20.60%=12(g)
⇒nFe2O3=\(\dfrac{12}{160}\)=0,075(mol
mCuO=20−12=8(g
⇒nCuO=\(\dfrac{8}{80}\)=0,1(mol)
Theo pT:
nFe=2nFe2O3=0,15(mol)
nCu=nCuO=0,1(mol)
⇒mFe=0,15.56=8,4(g)
mCu=0,1.64=6,4(g)
b, Theo PT: nH2=3nFe2O3+nCuO=0,325(mol)
⇒VH2=0,325.22,4=7,28(l)
c. Zn+2HCl->ZnCl2+H2
0,65----------0,325
=>m HCl=0,65.36,5=23,725g
\(nFeO=\dfrac{16}{56+16}=\dfrac{2}{9}\left(mol\right)\)
\(FeO+H_2\rightarrow Fe+H_2O\)
2/9 2/9 2/9 2/9
\(mFe=\dfrac{2}{9}.56=12,4\left(g\right)\)
\(VH_2=\dfrac{2}{9}.22,4=4,98\left(l\right)\)
a) Fe2O3 + 3 H2 → 2Fe + 3H2O
b) nFe2O3 = \(\dfrac{39}{160}\)= 0,24375 mol
=> nFe = 2nFe2O3 = 0,24375.2 = 0,4875 mol
c) Theo pt phản ứng nH2 = 3nFe
=> nH2 = 0,24375. 3 =0.73125 mol
<=> VH2 = 0.73125 . 22,4 = 16,38 lít
a, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2\left(mol\right)\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
c, \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\), ta được H2 dư.
Theo PT: \(n_{Cu}=n_{CuO}=0,2\left(mol\right)\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)
\(a/Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\\ b/n_{H_2}=n_{FeCl_2}=0,1mol\\ m_{FeCl_2}=0,1.127=12,7\left(g\right)\\ c/V_{H_2}=0,1.22,4=2,24\left(l\right)\\ d/n_{HCl}=0,1.2=0,2\left(mol\right)\\ V_{HCl\left(pư\right)}=\dfrac{0,2}{2}=0,1\left(l\right)\)
\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
Còn lại giống câu dưới nha
\(n_{FeO}=\dfrac{32}{72}=\dfrac{4}{9}\left(mol\right)\)
\(FeO+H_2\underrightarrow{^{t^0}}Fe+H_2O\)
\(\dfrac{4}{9}.....\dfrac{4}{9}....\dfrac{4}{9}\)
\(V_{H_2}=\dfrac{4}{9}\cdot22.4=10\left(l\right)\)
\(m_{Fe}=\dfrac{4}{9}\cdot56=24.89\left(g\right)\)
\(a.PTHH:3H_2+Fe_2O_3\rightarrow\left(t^o\right)2Fe+3H_2O\\b.n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\\ n_{Fe}=0,2.2=0,4\left(mol\right)\\ n_{H_2}=3.0,2=0,6\left(mol\right)\\ n_{Fe}=0,4.56=22,4\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,6.22,4=13,44\left(l\right) \)
a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
b, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe}=2n_{Fe_2O_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,4.56=22,4\left(g\right)\)
c, Theo PT: \(n_{H_2}=3n_{Fe_2O_3}=0,6\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)