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PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a_______a_______a_____a (mol)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
2b______3b__________b_____3b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}56a+27\cdot2b=11\\a+3b=0,2\cdot2=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{11}\cdot100\%\approx50,91\%\\\%m_{Al}=49,09\%\end{matrix}\right.\)
Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{FeSO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)
a) nH2SO4=0,4(mol)
Đặt: nFe=x(mol); nAl=y(mol) (x,y>0)
PTHH: Fe + H2SO4 -> FeSO4 + H2
x________x______x______x(mol)
2Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2
y____1,5y_______0,5y_______1,5y(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=> mFe=0,1.56=5,6(g)
=>%mFe=(5,6/11).100=50,909%
=>%mAl= 49,091%
b) V(H2,đktc)=0,4.22,4=8,96(l)
c) nAl2(SO4)3= 0,5y=0,5.0,2=0,1(mol)
nFeSO4=x=0,1(mol)
Vddsau=VddH2SO4=0,2(l)
=>CMddAl2(SO4)3= 0,1/0,2=0,5(M)
CMddFeSO4=0,1/0,2=0,5(M)
a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
X là khí Hidro
b) Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_{Al}=0,2mol\)
\(\Rightarrow\%m_{Al}=\dfrac{0,2\cdot27}{8,64}\cdot100\%=62,5\%\) \(\Rightarrow\%m_{Cu}=37,5\%\)
c) Theo PTHH: \(n_{HCl}=3n_{Al}=0,6mol\)
\(\Rightarrow V_{HCl}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)
Gọi x là số mol của Fe, R
Pt: Fe + 2HCl --> FeCl2 + H2
....0,05-> 0,1--------------> 0,05
.....R + 2HCl --> RCl2 + H2
0,05--> 0,1---------------> 0,05
Ta có: \(\left\{{}\begin{matrix}56x+xR=4\\127x+x\left(R+71\right)=11,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}56x+xR=4\\127x+xR+71x=11,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}56x+xR=4\\198x+xR=11,1\end{matrix}\right.\)
=> x = 0,05
=> R = 24
Vậy R là Magie (Mg)
VH2 = 0,05 . 2 . 22,4 = 2,24 (lít)
CM HCl = \(\dfrac{0,1+0,1}{0,1}=2M\)
mFe = 0,05 . 56 = 2,8 (g)
mMg = 4 - 2,8 = 1,2 (g)
% mFe = \(\dfrac{2,8}{4}.100\%=70\%\)
% mMg = 30%
\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)
Fe + H2SO4 → FeSO4 + H2
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Gọi x,y lần lượt là số mol Fe, Al
\(\left\{{}\begin{matrix}56x+27y=11\\x+\dfrac{3}{2}y=0,4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=>\(\%m_{Fe}=\dfrac{0,1.56}{11}.100=50,91\%\)
=> %m Al = 100 - 50,91 =49,09 %
b)Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\)
=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)
c) \(CM_{FeSO_4}=\dfrac{0,1}{0,2}=0,5M\)
\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{0,2}{2}}{0,2}=0,5M\)
2Al + 3H2SO4 => Al2(SO4)3 + 3H2
x 1,5x
Zn + H2SO4 => ZnSO4 + H2
y y
27x + 65y = 9,2
1,5x + y = 5,6/22,4
=> x= 0,1 y= 0,1
%Al = 29,348%
%Zn = 70,652%
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