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Mỗi phần a,b mol Fe , Al
\(n_S=n_{FeSO4}=n_{Fe}=a\left(mol\right)\)
\(n_{Al}=2n_{Al2\left(SO4\right)3}=b\left(mol\right)\)
\(\Rightarrow n_{Al2\left(SO4\right)3}=0,5b\left(mol\right)\)
\(\Rightarrow n_S=3n_{Al2\left(SO4\right)3}=1,5b\left(mol\right)\)
\(\%m_S=23,5204\%\Rightarrow\frac{32\left(a+1,5b\right)}{152a+342b}=0,235204\)
\(\Leftrightarrow-3,751008a-32,439768b=0\left(1\right)\)
\(n_{Br2}=1,245\left(mol\right)\)
\(SO_2+Br_2+2H_2O\rightarrow2HBr+H_2SO_4\)
1,245___1,245________________________
Bảo toàn e: \(3a+3b=1,245.2=2,49\left(2\right)\)
\(\left(1\right)+\left(2\right)\Rightarrow\) Nghiệm âm
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
a_______________________a
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b________________________a
\(\Rightarrow a+b=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(2Fe+3Cl_2\rightarrow2FeCl_3\)
a______1,5a_________
\(Mg+Cl_2\rightarrow MgCl_2\)
b_______b________
\(\Rightarrow1,5a+b=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=0,2\\1,5a+b=0,25\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\%m_{Mg}=\frac{0,1.24}{0,1.56+0,1.24}.100\%=30\%\)
Pt tác dụng H2SO4 loãng
CuO + H2SO4 \(\rightarrow\) CuSO4 + H2O (1)
Cu không tác dụng.
Cu + 2H2SO4đặc,n \(\rightarrow\) CuSO4 + SO2 + 2H2O (2)
nSO2= \(\frac{1,12}{22,4}\) = 0,05 mol
\(\rightarrow\) nCu= nSO2= 0,05 mol
% Cu = \(\frac{0,05x64}{10}.100\%\)= 32%
\(\rightarrow\) % CuO = 68%.
Câu 1:
Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
a________2a_______a______a(mol)
MgO +2 HCl -> MgCl2 + H2O
b_____2b_______b___b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> mMg=0,2.24=4,8(g)
=>%mMg= (4,8/8,8).100=54,545%
=> %mMgO= 45,455%
b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)
c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)
Câu 2:
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)
0,2____0,4_____0,2____0,2 (mol)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
0,2____0,4______0,2____0,2 (mol)
Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)
3/
a/
CuO+2HCl→CuCl\(_2\)+H\(_2\)O
0,05 0,1 0,05
2Al+6HCl→2AlCl\(_3\)+3H\(_2\)
0,1 0,3 0,1 0,15(mol)
n\(_{H_2}\)=\(\dfrac{3,36}{22,4}\)=0,15(mol)
m\(_{Al}\)=0,1.27=2,7(g)
⇔%m\(_{Al}\)=\(\dfrac{2,7}{6,7}.100\%\)≃40,3%
→%m\(_{CuO}\)=100%-40,3%=59,7%
b/
m\(_{CuO}\)=6,7-2,7=4(g)
n\(_{CuO}\)=\(\dfrac{4}{80}\)=0,05(mol)
m\(_{HCl}\)=(0,1+0,3).36,5=14,6(g)
m\(_{dd}\)=\(\dfrac{14,6.100\%}{14,6\%}\)=100(g)
m\(_{CuCl_2}\)=0,05.135=6,75(g)
C%\(_{CuCl_2}\)=\(\dfrac{6,75}{100}.100\%\)=6,75%
m\(_{AlCl_3}\)=0,1.133,5=13,35(g)
C%\(_{AlCl_3}\)=\(\dfrac{13,35}{100}.100\%\)=13,35%
1/
2Al+6HCl→2AlCl\(_3\)+3H\(_2\)
a a 3/2a (mol)
Fe+2HCl→FeCl\(_2\)+H\(_2\)
b b b (mol)
n\(_{H_2}\)=\(\dfrac{8,96}{22,4}\)=0,4(mol)
a/
gọi số mol của Al là a;số mol của Fe là b,ta có hệ phương trình:
27a+56b=11
\(\dfrac{3}{2}\)a+b=0,4
⇔ a=0,2(mol)
b=0,1(mol)
→m\(_{Al}\)=27.0,2=5,4(g)
⇔%m\(_{Al}\)=\(\dfrac{5,4}{11}.100\%\)≃49,1%
→%m\(_{Fe}\)=100%-49,1%=50,9%
b/
m\(_{AlCl_3}\)=0,2.133,5=26,7(g)
m\(_{FeCl_2}\)=0,1.127=12,7(g)
→m\(_{muối}\)=26,7+12,7=39,4(g)