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\(M+CuSO_4\rightarrow MSO_4+Cu\)
\(M+2AgNO_3\rightarrow M\left(NO_3\right)_2+2Ag\)
Lượng M phản ứng ở 2 PT trên là như nhau.
=> \(m_M=0,52-0,24=0,28\left(g\right)\)
Gọi x là nM, theo tăng giảm khối lượng có: \(108.2.x-64x-0,24=0,52\Rightarrow x=0,005\)
=> \(M_M=\dfrac{0,28}{0,005}=56\left(\dfrac{g}{mol}\right)\)
Vậy M là kim loại `Fe`
a) Gọi số mol Al, Fe là a, b (mol)
=> 27a + 56b = 11,1 (1)
\(n_{HCl}=\dfrac{60.36,5\%}{36,5}=0,6\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a--->3a-------->a----->1,5a
Fe + 2HCl --> FeCl2 + H2
b--->2b------->b----->b
=> 3a + 2b = 0,6 (2)
(1)(2) => a = 0,1; b = 0,15
\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{11,1}.100\%=24,32\%\\\%m_{Fe}=\dfrac{0,15.56}{11,1}.100\%=75,68\%\end{matrix}\right.\)
b) \(n_{H_2}=1,5a+b=\) 0,3 (mol)
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
mdd sau pư = 11,1 + 60 - 0,3.2 = 70,5 (g)
\(\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,1.133,5}{70,5}.100\%=18,94\%\\C\%_{FeCl_2}=\dfrac{0,15.127}{70,5}.100\%=27,02\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(m_{HCl}=\dfrac{60\cdot36,5}{100}=21,9g\)
\(\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x 3x 1,5x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y 2y y
\(\Rightarrow\left\{{}\begin{matrix}27x+56y=11,1\\3x+2y=0,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
a)\(\%m_{Al}=\dfrac{0,1\cdot27}{11,1}\cdot100\%=24,32\%\)
\(\%m_{Fe}=100\%-24,32\%=75,68\%\)
b)\(\Sigma n_{H_2}=1,5x+y=1,5\cdot0,1+0,15=0,3mol\)
\(V_{H_2}=0,3\cdot22,4=6,72l\)
\(a,Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ m_{tăng}=m_{hhMg,Al}-m_{H_2}\\ \Leftrightarrow7=7,8-m_{H_2}\\ \Leftrightarrow m_{H_2}=0,8\left(g\right)\\ \Rightarrow n_{H_2}=\dfrac{0,8}{2}=0,4\left(mol\right)\\ Đặt:a=n_{Al}\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}24b+27a=7,8\\b+1,5a=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\m_{Mg}=24.0,1=2,4\left(g\right);m_{Al}=0,2.27=5,4\left(g\right)\\ \Rightarrow\%m_{Al}=\dfrac{0,2.27}{7,8}.100\approx69,231\%\\ \Rightarrow\%m_{Mg}\approx30,769\%\\ c,m_{muối}=m_{MgSO_4}+m_{Al_2\left(SO_4\right)_3}=120b+342.0,5a=120.0,1+342.0,5.0,2=46,2\left(g\right)\)
\(n_{SO_2}=\dfrac{12,32}{22,4}=0,55mol\)
\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O+3SO_2\uparrow\)
x 3x 0,5x 3x 1,5x
\(2Ag+2H_2SO_4\rightarrow2H_2O+SO_2\uparrow+Ag_2SO_4\)
y y y 0,5y 0,5y
\(\Rightarrow\left\{{}\begin{matrix}1,5x+0,5y=0,55\\56x+108y=38,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
a)\(\%m_{Fe}=\dfrac{0,3\cdot56}{38,4}\cdot100\%=43,75\%\)
\(\%m_{Ag}=100\%-43,75\%=56,25\%\)
b)\(m_{muối}=m_{Fe_2\left(SO_4\right)_3}+m_{Ag_2SO_4}\)
\(\Rightarrow muối=0,5\cdot0,3\cdot400+0,5\cdot0,2\cdot312=91,2g\)
c)Cho hỗn hợp trên tác dụng \(H_2SO_4\) loãng chỉ có Fe tác dụng.
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,3 0,3 0,3
\(C_{M_{H_2SO_4}}=\dfrac{0,3}{0,2}=1,5M\)
\(V_{H_2}=0,3\cdot22,4=6,72l\)
\(n_{H_2}=\dfrac{2,464}{22,4}=0,11mol\)
\(\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\Rightarrow Muối\left\{{}\begin{matrix}Al_2\left(SO_4\right)_3\\FeSO_4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}BTe:3x+2y=2n_{H_2}=0,22\\\dfrac{x}{2}\cdot342+y\cdot152=14,44\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04mol\\y=0,05mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,04\cdot27=1,08g\\m_{Fe}=0,05\cdot56=2,8g\end{matrix}\right.\)
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\downarrow\)
0,02 0,06
\(FeSO_4+BaCl_2\rightarrow BaSO_4\downarrow+FeCl_2\)
0,05 0,05
\(\Rightarrow\Sigma n_{\downarrow}=0,06+0,05=0,11\Rightarrow m_{BaSO_4}=x=25,63g\)
Khối lượng đinh sắt tăng 2g => số mol Fe pư = 2: (64-56) = 0,25 mol