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\(a,n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,075<-------------------------0,075
Cu không phản ứng với H2SO4 loãng
b, \(m_{Mg}=0,075.24=1,8\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{1,8}{8}.100\%=22,5\%\\\%m_{Cu}=100\%-22,5\%=77,5\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\\ pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,075 0,075
\(Cu+H_2SO_4-x->\)
\(m_{Mg}=0,075.24=1,6\left(g\right)\\ m_{Cu}=8-1,6=6,4\left(g\right)\)
\(\%m_{Cu}=\dfrac{6,4}{8}.100\%=80\%\\
\%m_{Mg}=100-80\%=20\%\)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\)
\(b,\) Đặt \(n_{Fe}=x(mol);n_{Al}=y(mol)\)
\(\Rightarrow 56x+27y=8,3(1)\)
Theo PTHH: \(x+1,5y=0,25(2)\)
\((1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8,3}.100\%=67,47\%\\ \%_{Al}=100\%-67,47\%=32,53\%\)
\(a.2Al+6HCl->2AlCl_3+3H_2\\ Mg+2HCl->MgCl_2+H_2\\ b.n_{Al}=a,n_{Mg}=b\\ 27a+24b=7,5\left(I\right)\\ 1,5a+b=\dfrac{7,84}{22,4}=0,35\left(II\right)\\ a=0,1;b=0,2\\ \%m_{Al}=\dfrac{27\cdot0,1}{7,5}\cdot100\%=36\%\\ \%m_{Mg}=64\%\\ c.m_{HCl}=36,5\left(0,1\cdot3+0,2\cdot2\right)=18,25g\\ d.m_{ddsau}=7,5+\dfrac{18,25}{14,6:100}-0,35\cdot2=131,8g\\ C\%\left(AlCl_3\right)=\dfrac{133,5\cdot0,1}{131,8}\cdot100\%=10,1\%\\ C\%\left(MgCl_2\right)=\dfrac{95\cdot0,2}{131,8}\cdot100\%=14,4\%\)
Đặt \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\) ( mol ) \(\rightarrow m_{hh}=27x+56y=5,54\left(g\right)\) (1)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
x 1,5x ( mol )
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
y y ( mol )
\(n_{H_2}=1,5x+y=\dfrac{3,584}{22,4}=0,16\left(mol\right)\) (1)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,06\\y=0,07\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,06.27}{5,54}.100=29,24\%\\\%m_{Fe}=100-29,24=70,76\%\end{matrix}\right.\)
a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) Gọi x,y là số mol Al, Fe
\(n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Ta có hệ : \(\left\{{}\begin{matrix}27x+56y=0,83\\\dfrac{3}{2}x+y=0,02\end{matrix}\right.\)
=> \(x=\dfrac{29}{5700};y=\dfrac{47}{3800}\)
\(\%m_{Al}=\dfrac{\dfrac{27}{5700}.27}{0,83}.100=16,55\%\); \(\%m_{Fe}=100-16,55=83,45\%\)
c)Bảo toàn nguyên tố H: \(n_{H_2SO_4}=n_{H_2}=0,02\left(mol\right)\)
=> \(C\%_{H_2SO_4}=\dfrac{0,02.98}{200}.100=0,98\%\)
a)
Mg + 2HCl --> MgCl2 + H2
b)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,3<-----------------0,3
=> mMg = 0,3.24 = 7,2 (g)
=> mAg = 10,4 - 7,2 = 3,2 (g)
c) \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{10,4}.100\%=69,23\%\\\%m_{Ag}=\dfrac{3,2}{10,4}.100\%=30,77\%\end{matrix}\right.\)
Gọi $n_{Fe} = a(mol), n_{Zn} = b(mol) , n_{Al} = c(mol) \Rightarrow 56a + 65b + 27c = 20,4(1)$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$2Al +3 H_2SO_4 \to Al_2(SO_4)_3 +3 H_2$
Theo PTHH : $n_{H_2} = a + b + 1,5c = \dfrac{10,08}{22,4} = 0,45(mol)(2)$
Mặt khác :
$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$Zn + Cl_2 \xrightarrow{t^o} ZnCl_2$
$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
Theo PTHH : $n_{Cl_2} = 1,5n_{Fe} + n_{Zn} + 1,5n_{Al}$
Suy ra : \dfrac{1,5a + b + 1,5c}{a + b + c} = \dfrac{0,275}{0,2}(3)$
Từ (1)(2)(3) suy ra : a = 0,2 ; b = 0,1 ; c = 0,1
$\%m_{Fe} = \dfrac{0,2.56}{20,4}.100\% = 54,9\%$
$\%m_{Zn} = \dfrac{0,1.65}{20,4}.100\% = 31,9\%$
$\%m_{Al} = 100\% - 54,9\% - 31,9\% = 13,2\%$
a) PTHH: Mg + H2SO4 loãng \(\rightarrow\) MgSO4 + H2 \(\uparrow\)
2Al + 3H2SO4 loãng \(\rightarrow\) Al2(SO4)3 + 3H2 \(\uparrow\)
b) Gọi x,y lần lượt là số mol của Mg,Al có trong 6,3g hh ( x,y > 0 )
n\(H_2\) = \(\frac{6,72}{22,4}=0,3\left(mol\right)\)
Theo đề, ta có: mMg + mAl = 6,3
=> 24x + 27y = 6,3 (*)
Theo PT(1) : n\(H_2\) = nMg = x (mol)
Theo PT(2): n\(H_2\) = \(\frac{3}{2}\)nAl = \(\frac{3y}{2}\)(mol)
Vì n\(H_2\) =0,3(mol) => n\(H_2\)(1) + n\(H_2\)(2) = 0,3 (mol)
=> x + \(\frac{3}{2}y\) = 0,3 (**)
Từ (*) và (**) => \(\left\{{}\begin{matrix}24x+27y=6,3\\x+\frac{3}{2}y=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
=> mMg = 24.0,15 =3,6 (g)
Vậy thành phần % theo m mỗi kim loại trong hhA lần lượt là:
%mMg = \(\frac{3,6}{6,3}.100\%\approx57,143\%\)
%mAl = 100% - 57,143% =42,857%
+PTHH
Mg + H2SO4 => MgSO4 + H2
2Al + 3H2SO4 => Al2(SO4)3 + 3H2
nH2 = V/22.4 = 6.72/22.4 = 0.3 (mol)
Gọi x (mol), y (mol) lần lượt là số mol của Mg và Al
Ta có: \(\left\{{}\begin{matrix}24x+27y=6.3\\x+1.5y=0.3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0.15\\y=0.1\end{matrix}\right.\)
mMg = n.M = 0.15 x 24 = 3.6 (g)
mAl = n.M = 0.1 x 27 = 2.7 (g)
===> %mMg = 57.14 (%), %mAl = 42.86 (%)