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5)nH2SO4 b.đ=0,075(mol)
nH2SO4(pu2)=0,015
=>nH2SO4(p/u1)=0,06(mol)
X+H2SO4--->XSO4+H2
0,06----0,06
=>X=3,9/0,06=65(Zn)
1)Fe:a
Al:b
Hệ: \(\left\{{}\begin{matrix}56a+27b=8,3\\3a+3b=0,6\left(bte\right)\end{matrix}\right.\)
=>a=b=0,1
Đến đây thì zễ ùi
Ta có:
\(n_{SO2}=\frac{2,668}{22,4}=0,12\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Zn}:x\left(mol\right)\\n_{ZnO}:y\left(mol\right)\end{matrix}\right.\)
\(Zn+2H_2SO_4\rightarrow ZnSO_4+SO_2+2H_2O\)
x_____2x__________________x______
\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
y_______y____________________
\(\Rightarrow x=0,12\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,12.65=7,8\left(g\right)\\m_{ZnO}=27,24-7,8=19,44\left(g\right)\end{matrix}\right.\)
\(n_{ZnO}=\frac{19,44}{81}=0,24\left(mol\right)\)
\(n_{H2SO4}=0,48\left(mol\right)\)
\(\Rightarrow m_{H2SO4}=0,48.96=46,08\left(g\right);m_{dd\left(H2SO4\right)}=\frac{46,08}{80\%}=57,6\left(g\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a_____a_________________ a
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b_____ b ___________________b
Giải hệ PT:
\(\left\{{}\begin{matrix}56a+65b=12,1\\a+b=\frac{4,48}{22,4}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{0,1.56}{12,1}.100\%=46,28\%\\\%m_{Zn}=100\%-46,28\%=53,72\%\end{matrix}\right.\)
\(\Rightarrow V_{H2SO4\left(can.dung\right)}=\frac{0,1+0,1}{0,2}=0,1\left(l\right)\)
\(\Rightarrow\left\{{}\begin{matrix}CM_{FeSO4}=\frac{0,1}{0,1}=1M\\CM_{ZnSO4}=\frac{0,1}{0,1}=1M\end{matrix}\right.\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
a a a a (mol)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
b \(\dfrac{3}{2}b\) \(\dfrac{1}{2}b\) \(\dfrac{3}{2}b\) (mol)
n\(_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
gọi số mol của Mg trong hỗn hợp là a;Al là b,ta có hệ phương trình:
\(\left\{{}\begin{matrix}a+\dfrac{3}{2}b=0,4\\24a+27b=7,8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
m\(_{Mg}=0,1.24=2,4\left(g\right)\)
\(\rightarrow m_{Al}=7,8-2,4=5,4\left(g\right)\)
b/
\(V_{H_2SO_4}=\dfrac{0,1+0,3}{2}=0,2\left(l\right)=200\left(ml\right)\)
2/
\(Cu+2H_2SO_{4\left(đn\right)}\rightarrow CuSO_4+SO_2+2H_2O\)
0,2 0,2 (mol)
\(n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)
m\(_{ddNaOH}=1,28.125=160\left(g\right)\)
\(m_{NaOH}=\dfrac{160.25}{100}=40\left(g\right)\)
\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)
\(T=\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{1}{0,2}=5\)
vậy phản ứng sẽ tạo ra muối natri sulfit(\(Na_2SO_3\)) và dư NaOH:
\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)
bđ: 1 0,2 0 (mol)
pư: 0,4 0,2 0,2 (mol)
dư: 0,6 0 0 (mol)
\(C_{M_{Na_2SO_3}}=\dfrac{0,2}{0,125}=1,6\left(M\right)\)
Fe+H2SO4->FeSO4+H2
0,1-----------------------0,1
Fe2O3+3H2SO4->Fe2(SO4)3+3H2O
0,1-------------------0,1
nH2=2,24\22,4=0,1 mol
=>mFe=0,1.56=5,6g
=>mFe2O3=16g=>nFe2O3=0,1 mol
m muối =0,1.152+0,1.400=55,2g
