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\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
(mol).......0,3........0,6.........0,3.......0,3
a) \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b) \(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
c) \(200ml=0,2l\)
\(C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\left(M\right)\)
d) \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
\(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Ban đầu: 0,2......0,3
Phản ứng: 0,2....0,2.....0,2.....0,2
Dư:.....................0,1
Lập tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\left(0,2< 0,3\right)\)
\(\Rightarrow H_2\) dư
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
PTHH :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,3 0,6 0,3 0,3
\(a,V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(b,C_{M_{HCl}}=\dfrac{n}{V}=\dfrac{0,6}{0,5}=1,2M\)
\(c,m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
Tên gọi : Kẽm Clorua
a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1--->0,2------->0,1----->0,1
VH2 = 0,1.22,4 = 2,24 (l)
b, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)
c, \(C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,4
0,1 0,2
0 0,2 0,1 0,1
\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\\ b,m_{dd}=6,5+0,4.36,5+50-0,1.2=70,9\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{70,9}.100\%=19,18\%\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,1-->0,2
=> \(C_{M\left(dd.HCl\right)}=\dfrac{0,2}{0,2}=1M\)
`n_(Zn) = (6,5)/65 = 0,1 (mol)`
`PTHH: ZN + 2HCl --> ZnCl_2 + H_2`
`0,1 --> 0,2`
`=> C_(M(dd.HCl) = (0,2)/(0,2) = 1M`
\(n_{Fe}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1 0,2 0,1 0,1
\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(b,m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C\%=\dfrac{7,3}{73}.100\%=10\%\)
\(c,m_{ddZnCl_2}=6,5+73-\left(0,1.2\right)79,3\left(g\right)\)
\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
\(C\%_{ZnCl_2}=\dfrac{13,6}{79,3}.100\%=17,15\%\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1 2
Vì 0,1/1<2/2
nên tính theo Zn
=>\(n_{H_2}=n_{Zn}=0.1\left(mol\right)\) và nHCl=0,2(mol)
\(V=0.1\cdot22.4=2.24\left(lít\right)\)
\(C\%\left(muối\right)=\dfrac{0.1\cdot136}{6.5+73-0.2}\simeq17,15\%\)