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Tính %m mỗi oxit chứ:v
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe_3O_4}=x\left(mol\right)\\n_{ZnO}=y\left(mol\right)\end{matrix}\right.\)
\(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
x --------> 4x ---> 3x
\(ZnO+H_2\underrightarrow{t^o}Zn+H_2O\)
y ------> y --> y
Có hệ phương trình \(\left\{{}\begin{matrix}232x+81y=19,7\\4x+y=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
\(\%_{m_{Fe_3O_4}}=\dfrac{232.0,05.100}{19,7}=58,88\%\)
\(\%_{m_{ZnO}}=\dfrac{81.0,1.100}{19,7}=41,12\%\)
\(n_{Fe}=3x=3.0,05=0,15\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\\ n_{Zn}=y=0,1\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
a) \(n_{H_2}:\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Gọi x, y lần lượt là số mol của \(Fe_3O_4,ZnO\)
\(Fe_3O_4+4H_2\rightarrow3Fe+4H_2O\)
1...................4............3............4(mol)
x..................4x.........3x...........4x(mol)
\(ZnO+H_2\rightarrow Zn+H_2O\)
1..............1...........1.........1(mol)
y..............y............y.........y(mol)
Ta có:\(\left\{{}\begin{matrix}232x+81y=19,7\\4x+y=0,3\end{matrix}\right.\)
=>x=0,05
=>y=0.1
\(m_{Fe_3O_4}:232.0,05=11,6\left(g\right)\)
\(m_{ZnO}:19,7-11,6=8,1\left(g\right)\)
b)\(m_{Fe}:56.0,15=8,4\left(g\right)\)
\(m_{Zn}:65.0,1=6,5\left(g\right)\)
c)\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
....1................1..................1............1(mol)
0,3................0,3................0,3.........0,3(mol)
\(m_{Mg}:0,3.24=7,2\left(g\right)\)
\(m_{H_2SO_4}:0,3.98+0,3.98.10\%=32.34\left(g\right)\)
cumg co mot cach lam khac cua cau a
do la bang cach goi x la so mol cua H2 tham gia vao pthh 1 Fe3O4+4H2 \(\rightarrow\)3Fe+4H2O
so mol cua H2 tham gia vao pthh 2 la 0,3-x mol
a) CuO + H2 -> Cu + H2O (1)
ZnO + H2 -> Zn + H2O (2)
n\(_{H_2}\) = \(\dfrac{4,48}{22,4}\) = 0,2 (mol) => m\(H_2\) = 0,2.2 = 0,4 (g)
Theo PT (1) và (2) ta có: n\(H_2O\) = n\(H_2\) = 0,2 (mol)
=> m\(H_2O\) = 0,2.18 = 3,6 (g)
Theo định luật bảo toàn khối lượng ta có:
mhh + m\(H_2\) = mkim loại + m\(H_2O\)
=> mhh = 12,9 + 3,6 - 0,4 = 16,1 (g)
b) Gọi mCu là x(g) (0<x<12,9) => nCu = \(\dfrac{x}{64}\) (mol)
Thì mZn là 12,9-x (g) => nZn = \(\dfrac{12,9-x}{65}\) (mol)
Theo PT (1) ta có: n\(H_2\) = nCu = \(\dfrac{x}{64}\) (mol)
Theo PT (2) ta có: n\(H_2\) = nZn = \(\dfrac{12,9-x}{65}\) (mol)
Theo đề bài, n\(H_2\) là 0,2mol nên ta có:
\(\dfrac{x}{64}+\dfrac{12,9-x}{65}=0,2\) <=> x = 6,4 (g)
=> mCu = 6,4 (g)
Vậy: %Cu = \(\dfrac{6,4}{12,9}\).100% \(\approx\) 49,61%
%Zn = 100% - 49,61% = 50,39%
ta co pthh
Fe3O4+4 H2 \(\rightarrow\)3Fe +4 H2O
ZnO+H2 \(\rightarrow\)Zn + H2O
Theo de bai ta co nH2= \(\dfrac{6,72}{22,4}=0,3mol\)
goi x la so mol cua H2 tham gia vao pthh 1
so mol cua H2 tham gia vao pthh 2 la 0,3-x mol
theo pthh 1 nFe3O4= \(\dfrac{1}{4}nH2=\dfrac{1}{4}x\) mol
theo pthh 2 nZnO=nH2= 0,3-x mol
theo de bai ta co
232.\(\dfrac{1}{4}x\)+ 81.(0,3-x)=19,7
\(\Leftrightarrow\)58x + 24,3 -81x = 19,7
\(\Leftrightarrow\)-23x=19,7-24,3
\(\Leftrightarrow\)-23x=-4,6
\(\Rightarrow\)x= \(\dfrac{-4,6}{-23}=0,2mol\)
\(\Rightarrow\)nFe3O4=\(\dfrac{1}{4}nH2=\dfrac{1}{4}.0,2=0,05mol\)
nZnO=nH2=0,3-0,2=0,1 mol
\(\Rightarrow\)Khoi luong moi oxit trong hh la
mFe3O4=232.0,05=11,6 g
mZnO= mhh-mFe3O4=19,7-11,6=8,1 g
Theo pthh1 nFe= \(\dfrac{3}{4}nH2=\dfrac{3}{4}.0,2=0,15mol\)
\(\Rightarrow\)mFe= 0,15.56=8,4 g
theo pthh 2 nZn=nH2= 0,1 mol
\(\Rightarrow\)mZn=0,1.65=6,5 g
b6
nH2=V/22,4=5,04/22,4=0,225(mol)
Gọi a,b lần lượt là sô mol của Fe2O3 và CuO
pt1: Fe2O3 + 3H2-t0-> 2Fe +3H2O
cứ::1.................3..........2.............3 (mol)
vậy: a----------->3a------>2a (mol)
pt2: CuO +H2 -t0-> Cu +H2O
cứ:: 1...........1.............1........1 (mol)
vậy: b--------->b-------->b (mol)
từ 2pt và đề ta có:
160a+80b=14
3a+b=0,225
=> a=0,05(mol) ;b=0,075(mol)
=> mFe=n.M=0,05.56=2,8(g)
mCu=n.M=0,075.64=4,8(g)
=> mhh hai kim loại= mFe +mCu=2,8+4,8=7,6(g)
c) Pt3: Zn +2HCl -> ZnCl2 +H2
cứ::;; 1............2...........1..........1 (mol)
vậy: 0,225<---0,45<---0,225<--0,225(mol)
=> mZn=n.M=0,225.65=14,625(g)
mHCl=n.M=0,45.36,5=16,425(g)
b7
a) nH2:6,7222,4=0,3(mol)
Gọi x, y lần lượt là số mol của Fe3O4,ZnO
Fe3O4+4H2→3Fe+4H2O
1...................4............3............4(mol)
x..................4x.........3x...........4x(mol)
ZnO+H2→Zn+H2O
1..............1...........1.........1(mol)
y..............y............y.........y(mol)
Ta có:
{232x+81y=19,74
x+y=0,3
=>x=0,05
=>y=0.1
mFe3O4:232.0,05=11,6(g
mZnO:19,7−11,6=8,1(g)
b)mFe:56.0,15=8,4(g)
mZn:65.0,1=6,5(g)
c)Mg+H2SO4→MgSO4+H2
....1................1..................1............1(mol)
0,3................0,3................0,3.........0,3(mol)
mMg:0,3.24=7,2(g)mMg:0,3.24=7,2(g)
mH2SO4:0,3.98+0,3.98.10%=32.34(g)
Bài 1:
a) PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+2H_2O\)
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
b) Ta có: \(\left\{{}\begin{matrix}m_{Fe_2O_3}=20\cdot80\%=16\left(g\right)\\m_{CuO}=20-16=4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2}=3n_{Fe_2O_3}+n_{CuO}=0,35\left(mol\right)\) \(\Rightarrow V_{H_2}=0,35\cdot22,4=7,84\left(l\right)\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\\n_{Cu}=n_{CuO}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{hhB}=m_{Fe}+m_{Cu}=0,2\cdot56+0,05\cdot64=14,4\left(g\right)\)
Bài 2:
PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
a) Vì khối lượng Cu bằng \(\dfrac{6}{5}\) khối lượng Fe
\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=\dfrac{26,4}{6+5}\cdot6=14,4\left(g\right)\\m_{Fe}=26,4-14,4=12\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=\dfrac{14,4}{64}=0,225\left(mol\right)\\n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2}=\dfrac{3}{2}n_{Fe}+n_{Cu}=\dfrac{9}{28}+0,225=\dfrac{153}{280}\left(mol\right)\) \(\Rightarrow V_{H_2}=\dfrac{153}{280}\cdot22,4=12,24\left(l\right)\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{3}{28}\left(mol\right)\\n_{CuO}=n_{Cu}=0,225\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=\dfrac{3}{28}\cdot160\approx17,14\left(g\right)\\m_{CuO}=0,225\cdot80=18\left(g\right)\end{matrix}\right.\) \(\Rightarrow m_{hh}=35,14\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{17,14}{35,14}\cdot100\%\approx48,78\%\\\%m_{CuO}=51,22\%\end{matrix}\right.\)
Ta có nH2 = \(\dfrac{6,72}{22,4}\) = 0,3 ( mol )
Fe3O4 + 4H2 \(\rightarrow\) 3Fe + 4H2O
x................4x.......3x.........4x
ZnO + H2 \(\rightarrow\) Zn + H2O
y...........y.........y........y
=> \(\left\{{}\begin{matrix}232x+81y=19,7\\4x+y=0,3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
a, => mFe3O4 = 232 . 0,05 = 11,6 ( gam )
=> mZnO = 81 . 0,1 = 8,1 ( gam )
b, => mFe = 56 . ( 0,05 . 3 ) = 8,4 ( gam )
=> mZn = 65 . 0,1 = 6,5 ( gam )
c,
Mg + H2SO4 \(\rightarrow\) MgSO4 + H2
0,3........0,3............0,3.........0,3
=> mMg = 0,3 . 24 = 7,2 ( gam )
=> mH2SO4 = 98 . 0,3 = 29,4 ( gam )
=> mH2SO4 cần dùng = 29,4 : 90 . 100 = \(\dfrac{49}{15}\) ( gam )