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\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,05 0,15
\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)
\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)
a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)
\(\%m_{Ag}=100\%-64,28\%=35,72\%\)
b)\(m_{muối}=0,05\cdot342=17,1g\)
a) Gọi số mol Al, Mg là a, b
=> 27a + 24b = 6,3
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
a------------------------->1,5a
Mg + 2HCl --> MgCl2 + H2
b--------------------------->b
=> \(1,5a+b=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
=> a = 0,1; b = 0,15
=> \(\left\{{}\begin{matrix}m_{Al}=0,1.27=2,7\left(g\right)\\m_{Mg}=0,15.24=3,6\left(g\right)\end{matrix}\right.\)
b)
PTHH: MxOy + yH2 --to--> xM + yH2O
\(\dfrac{0,3}{y}\)<--0,3
=> \(M_{M_xO_y}=x.M_M+16y=\dfrac{17,4}{\dfrac{0,3}{y}}\)
=> \(M_M=21.\dfrac{2y}{x}\left(g/mol\right)\)
Xét \(\dfrac{2y}{x}=1\) => Loại
Xét \(\dfrac{2y}{x}=2\) => Loại
Xét \(\dfrac{2y}{x}=3\) => Loại
Xét \(\dfrac{2y}{x}=\dfrac{8}{3}\) => MM = 56 (g/mol) => M là Fe
a, ptpứ:
\(Mg+2HCl\rightarrow MgCl_2+H_2\left(1\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(2\right)\)
gọi số mol Mg là x mol , số mol Al là y mol ( x; y >0)
ta có pt : \(24x+27y=6,3\left(3\right)\)
theo bài : \(nH_2=0,3mol\)
theo ptpư(1) \(nH_2=nMg=xmol\)
theo ptpư(2) \(nH_2=\dfrac{3}{2}nAl=\dfrac{3}{2}ymol\)
tiếp tục có pt : \(x+\dfrac{3}{2}y=0,3\left(4\right)\)
từ (3) và (4) ta có hệ pt:
\(24x+27y=6,3\\ x+\dfrac{3}{2}y=0,3\)
<=> \(x=0,15\) ; \(y=0,1\)
\(mMg=24x=24.0,15=3,6gam\)
\(mAl=27y=27.0,1=2,7gam\)
a, Cu không tác dụng với dd HCl.
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)
Bạn tham khảo nhé!
\(a.2Al+6HCl->2AlCl_3+3H_2\\ Mg+2HCl->MgCl_2+H_2\\ b.n_{Al}=a,n_{Mg}=b\\ 27a+24b=7,5\left(I\right)\\ 1,5a+b=\dfrac{7,84}{22,4}=0,35\left(II\right)\\ a=0,1;b=0,2\\ \%m_{Al}=\dfrac{27\cdot0,1}{7,5}\cdot100\%=36\%\\ \%m_{Mg}=64\%\\ c.m_{HCl}=36,5\left(0,1\cdot3+0,2\cdot2\right)=18,25g\\ d.m_{ddsau}=7,5+\dfrac{18,25}{14,6:100}-0,35\cdot2=131,8g\\ C\%\left(AlCl_3\right)=\dfrac{133,5\cdot0,1}{131,8}\cdot100\%=10,1\%\\ C\%\left(MgCl_2\right)=\dfrac{95\cdot0,2}{131,8}\cdot100\%=14,4\%\)
2Al+6HCl→2AlCl3+3H2
Zn+2HCl→ZnCl2+H2
2Al+6H2SO4→Al2(SO4)3+3SO2+6H2O
Zn+2H2SO4→ZnSO4+SO2+2H2O
Cu+2H2SO4→CuSO4+SO2+2H2O
nH2=0,3mol
nCu=0,15mol
Gọi a và b lần lượt là số mol của Al và Zn
27a+65b=17,25
3\2a+b=0,3
=> a=0,03, b=0,25
→nAl=0,03mol→mAl=1,62g
→nZn=0,25mol→mZn=32,5g
b)nHCl=3nAl+2nZn=0,59mol
→VHCl=0,592=0,295 l
c)
nAl2(SO4)3=1\2nAl=0,015mol
→mAl2(SO4)3=5,13g
nZnSO4=nZn=0,25mol
→mZnSO4=40,25g
nCuSO4=nCu=0,15mol
→mCuSO4=24g
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ 2Al+6HCl\to 2AlCl_3+3H_2\)
\(b,\) Đặt \(n_{Fe}=x(mol);n_{Al}=y(mol)\)
\(\Rightarrow 56x+27y=8,3(1)\)
Theo PTHH: \(x+1,5y=0,25(2)\)
\((1)(2)\Rightarrow x=y=0,1(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,1.56}{8,3}.100\%=67,47\%\\ \%_{Al}=100\%-67,47\%=32,53\%\)
Bài 1:
\(n_{H_2SO_4}=\dfrac{300.19,6\%}{98}=0,6\left(mol\right);n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,6}{3}\Rightarrow H_2SO_4dư\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{3n_{Al}}{2}=\dfrac{3.0,1}{2}=0,15\left(mol\right)\\ a,m_{Al_2\left(SO_4\right)_3}=342.0,15=51,3\left(g\right)\\ b,m_{ddsau}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=2,7+300-\dfrac{3}{2}.0,1.2=302,4\left(g\right)\\ c,C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{51,3}{302,4}.100\%\approx16,964\%\\ n_{H_2SO_4\left(dư\right)}=0,6-\dfrac{3}{2}.0,1=0,45\left(mol\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{0,45.98}{302,4}.100\%\approx14,583\%\)
Bài 2:
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ Hpt:\left\{{}\begin{matrix}27a+24b=7,8\\1,5a+b=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \%m_{Al}=\dfrac{0,2.27}{7,8}.100\%\approx69,231\%\Rightarrow\%m_{Mg}\approx100\%-69,231\%\approx30,769\%\)