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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 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
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 )
Gọi nFe=a(mol);nM=b(mol)⇒56a+Mb=9,6(1)
Fe+2HCl→FeCl2+H2
M+2HCl→MCl2+H2
nH2=a+b=0,2⇒a=0,2−b
Ta có :
56a+Mb=9,656a+Mb=9,6
⇔56(0,2−b)+Mb=9,6
⇔Mb−56b=−1,6
⇔b(56−M)=1,6
⇔b=1,656−M
Mà 0<b<0,20<b<0,2
Suy ra : 0<1,656−M<0,20<1,656−M<0,2
⇔M<48(1)
M+2HCl→MCl2+H2
nM=nH2<5,622,4=0,25
⇒MM>4,60,25=18,4
+) Nếu M=24(Mg)
Ta có :
56a+24b=9,656a+24b=9,6
a+b=0,2a+b=0,2
Suy ra a = 0,15 ; b = 0,05
mFe=0,15.56=8,4(gam)
mMg=0,05.24=1,2(gam)
+) Nếu M=40(Ca)
56a+40b=9,656a+40b=9,6
a+b=0,2
Suy ra a = b = 0,1
mCa=0,1.40=4(gam)
mFe=0,1.56=5,6(gam)
\(n_{O_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(n_{H_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(n_{Fe}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(a.......\dfrac{2a}{3}\)
\(4Al+3O_2\underrightarrow{^{^{t^0}}}2Al_2O_3\)
\(b.......\dfrac{3b}{4}\)
\(n_{O_2}=\dfrac{2a}{3}+\dfrac{3b}{4}=0.25\left(mol\right)\left(1\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{H_2}=a+1.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.15,b=0.2\)
\(m_{Fe}=0.15\cdot56=8.4\left(g\right)\)
\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)
\(\%m_{Fe}=\dfrac{8.4}{8.4+5.4}\cdot100\%=60.8\%\)
\(\%m_{Al}=100-60.8=39.2\%\)
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)