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a) nAl=0,2(mol)
PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
H2 + CuO -to-> Cu + H2O
nAlCl3= nAl= 0,2(mol)
=> mAlCl3= 133,5. 0,2= 26,7(g)
b) nCu= nH2= 3/2 . 0,2=0,3(mol)
=> mCu= 0,3.64=19,2(g)
(Qua phản ứng nghe kì á, chắc tạo thành chứ ha)
<3
\(a,2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{HCl}=0,2.1,5=0,6\left(mol\right)\\ n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right);n_{Al}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ b,V=V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ m=m_{Al}=0,2.27=5,4\left(g\right)\)
\(a,n_{MgCO_3}=\dfrac{8,4}{84}=0,1\left(mol\right)\)
PTHH: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\)
0,1------->0,2-------->0,1-------->0,1
\(\rightarrow\left\{{}\begin{matrix}V=0,1.22,4=2,24\left(l\right)\\a=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\\ b,m_{muối}=0,1.95=9,5\left(g\right)\)
nH2 = 6,72 : 22,4 = 0,3 ( mol )
PTHH : 2Al + 6HCl -> 2AlCl3 + 3H2
0,2 0,6 mol <- 0,3mol
a = mAl = 0,2 x 27 = 5,4 (g)
VHCl = 0,6 : 2 = 0,3 ( l ) = 300 ( ml )
\(a)Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b)n_{Mg}=\dfrac{3}{24}=0,125mol\\ n_{HCl}=0,1.1=0,1mol\\ \Rightarrow\dfrac{0,125}{1}>\dfrac{0,1}{2}\Rightarrow Mg.dư\\ n_{H_2}=n_{MgCl_2}=\dfrac{0,1}{2}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5M\)
Đoạn xét tỉ lệ phải là \(\dfrac{0,125}{1}>\dfrac{0,1}{2}\) em nhé.
\(a\)) \(PTHH:Zn+2HCl\underrightarrow{t^o}ZnCl_2+H_2\)
0,25 0,5 0,25 0,25
b) nZn=\(\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\)
\(V_{H_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)
c) \(m_{HCl}=n.M=0,5.36,5=18,25\left(g\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{18,48}{22,4}=0,825\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}24x+27y=17,1\\x+\dfrac{3}{2}y=0.825\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,3\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,375.24}{17,1}.100=52,63\%\\ \%m_{Al}=47,37\%\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ a,PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{HCl}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ b,V=V_{ddHCl}=\dfrac{0,3}{3}=0,1\left(l\right)=100\left(ml\right)\)