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PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
a+b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,2\left(mol\right)\\n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,2\cdot160=32\left(g\right)\\V_{H_2}=0,6\cdot22,4=13,44\left(l\right)\end{matrix}\right.\)
c) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Theo PTHH: \(n_{Zn}=n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,6\cdot65=39\left(g\right)\)
a,
nFe = 22,4/56 = 0,4 (mol)
PTHH
Fe2O3 + 3H2 ---to----) 2Fe + 3H2O (1)
theo phương trình (1) ,ta có:
nFe2O3 = 0,4 x 2 / 1 = 0,8 (mol)
mFe2O3 = 160 x 0,8 = 128 (g)
b,
theo pt (1)
nH2 = (0,4 x 3)/2 = 0,6 (mol)
=) VH2 = 0,6 x 22,4 = 13,44 (L)
c,
PTHH
Zn + H2SO4 -------------) ZnSO4 + H2 (2)
Số mol H2 cần dùng là 0,6 (mol)
Theo PT (2) :
nZn = nH2 ==) nZn = 0,6 x 65 = 39 (g)
a) \(n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\dfrac{3}{14}\)---------------------->\(\dfrac{3}{14}\)
\(\Rightarrow V_{H_2}=\dfrac{3}{14}.22,4=4,8\left(l\right)\)
b) \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
PTHH: \(ZnO+H_2\xrightarrow[]{t^o}Zn+H_2O\)
Xét tỉ lệ: \(0,1< \dfrac{3}{14}\Rightarrow H_2\) dư
Theo PT: \(n_{Zn}=n_{ZnO}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(1mol\) \(1mol\)
\(\dfrac{3}{14}mol\) \(\dfrac{3}{14}mol\)
\(a)n_{Fe}=\dfrac{m}{M}=\dfrac{12}{56}\approx0,21=\dfrac{3}{14}\left(mol\right)\)
\(V_{H_2}=n.22,4=\dfrac{3}{14}.22,4=4,8\left(l\right)\)
\(b)n_{ZnO}=\dfrac{m}{M}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
\(ZnO+H_2\rightarrow Zn+H_2O\)
\(1mol\) \(1mol\) \(1mol\)
\(0,1mol\) \(0,1mol\) \(0,1mol\)
\(\text{Ta thấy }H_2\text{ dư,ZnO phản ứng hết.Bài toán tính theo ZnO}\)
\(m_{Zn}=n.M=0,1.65=6,5\left(g\right)\)
nFe2O3 = 16,8/56 = 0,3 (mol)
PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
MOL: 0,15 <--- 0,45 <--- 0,3
VH2 = 0,45 . 22,4 = 10,08 (l)
mFe2O3 = 0,45 . 160 = 72 (g)
a ) Fe2O3 + 3H2 ---> 2Fe + 3H2O
nFe = 16,8 :56 =0,3
Fe2O3 + 3H2--> 2Fe +3H2O
0,15<------0,45<---- 0,3
VH2 = 0,45.22,4=10,08(l)
mFe2O3 = 0,15.160 =24(g)
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
$1)$
$PTHH:Ca(OH)_2+2HCl\to CaCl_2+2H_2O$
$n_{Ca(OH)_2}=\dfrac{14,8}{74}=0,2(mol)$
$n_{HCl}={10,95}{36,5}=0,3(mol)$
Lập tỉ lệ: $\dfrac{n_{Ca(OH)_2}}{1}>\dfrac{n_{HCl}}{2}\Rightarrow Ca(OH)_2$ dư
$\Rightarrow n_{Ca(OH)_2(dư)}=0,2-\dfrac{1}{2}.0,3=0,05(mol)$
Theo PT: $n_{CaCl_2}=\dfrac{1}{2}n_{HCl}=0,15(mol)$
$\Rightarrow m_{CaCl_2}=0,15.111=16,65(g)$
$m_{Ca(OH)_2(dư)}=0,05.74=3,7(g)$
$2)$
$a)PTHH:Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\uparrow$
$b)n_{Fe}=\dfrac{22,4}{56}=0,4(mol)$
Theo PT: $n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,2(mol)$
$n_{CO}=\dfrac{3}{2}n_{Fe}=0,6(mol)$
$\Rightarrow m_{Fe_2O_3}=0,2.160=32(g)$
$V_{CO}=0,6.22,4=13,44(lít)$
\( n_{Fe} =\dfrac{11,2}{56} = 0,2(mol)\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ n_{H_2} = \dfrac{3}{2}n_{Fe} = 0,3(mol)\\ n_{HCl} = 2n_{H_2} = 0,3.2 = 0,6(mol)\\ \Rightarrow m_{HCl} = 0,6.36,5 = 21,9(gam)\)