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a/ \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,3 0,6 0,3 0,3
\(m_{Mg}=0,3.24=7,2\left(g\right)\)
b/ \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c/ \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,1\left(mol\right)\\n_{CuCl_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,1\cdot36,5}{7,3\%}=50\left(g\right)\\C\%_{CuCl_2}=\dfrac{0,05\cdot135}{4+50}\cdot100\%=12,5\%\end{matrix}\right.\)
a) Đặt: nZn=x(mol); nFe= y(mol) (x,y: nguyên, dương)
Zn + H2SO4 -> ZnSO4 + H2
x_______x_______x________x
Fe + H2SO4 -> FeSO4 + H2
y____y_________y___y(mol)
b) m(rắn)=mCu=3(g)
=> m(Zn, Fe)= 21,6 - 3= 18,6(g)
Ta có hpt:
\(\left\{{}\begin{matrix}65x+56y=18,6\\22,4x+22,4y=6,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
=> Zn= 65.0,2=13(g)
=>%mZn= (13/21,6).100=60,185%
%mCu=(3/21,6).100=13,889%
=>%mFe=25,926%
c) nH2SO4=x+y=0,3(mol) =>mH2SO4=29,4(g)
=> mddH2SO4= (29,4.100)/25=117,6(g)
a) \(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
0,5------->0,5------------------------>0,5
=> mC2H5OH = 0,5.46 = 23 (g)
b) mCH3COOC2H5 = 0,5.88 = 44 (g)
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{3,7185}{24,79}=0.,15(mol)\\ \Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{10}.100\%=97,5\%\\ \Rightarrow \%_{Cu}=100\%-97,5\%=2,5\%\\ b,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{14\%}=78,21(g)\)
Ta vó nH2 = \(\dfrac{13,44}{22,4}\) = 0,6 ( mol )
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
0,4.......0,6...........0,4........0,6
=> mAl = 27 . 0,4 = 10,8 ( gam )
=> %mAl = \(\dfrac{10,8}{27}\) . 100 = 40 %
=> %mZn = 100 - 40 = 60 %
=> mZnO = 27 - 10,8 = 16,2 ( gam )
=> nZnO = \(\dfrac{16,2}{81}\) = 0,2 ( mol )
ZnO + 2HCl \(\rightarrow\) ZnCl2 + H2O
0,2........0,4.........0,2
=> mHCl = ( 0,4 + 0,6 ) . 36,5 = 36,5 ( gam )
=> mHCl cần dùng = 36,5 : 29,2 . 100 = 125 ( gam )
=> mAlCl3 = 0,4 . 133,5 = 53,4 ( gam )
=> mZnCl2 = 136 . 0,2 = 27,2 ( gam )
Mdung dịch = Mtham gia - MH2
= 27 + 125 - 0,6 . 2
= 150,8 ( gam )
==> C%AlCl3 = \(\dfrac{53,4}{150,8}\times100\approx35,4\%\)
=> C%ZnCl2 = \(\dfrac{27,2}{150,8}\times100\approx18,04\%\)
\(PTHH:Cu+Cl_2\xrightarrow{t^o} CuCl_2\\ \Rightarrow n_{Cu}=n_{CuCl_2}=n_{Cl_2}=0,1(mol)\\ \Rightarrow m_{Cu}=0,1.64=6,4(g); m_{CuCl_2}=0,1.135=13,5(g)\)