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a, PTHH: Mg+2HCl--->MgCl2+H2
b, nHCl= \(\dfrac{10,95}{36,5}=0,3\) mol
Theo pt: nMg=\(\dfrac{1}{2}.nHCl=\dfrac{1}{2}.0,3=0,15\) mol
=> mMg= 0,15.24= 3,6 (g)
c, Theo pt: nH2=\(\dfrac{1}{2}.nHCl=\dfrac{1}{2}.0,3=0,15\) mol
=> VH2= 0,15.22,4= 3,36 (l)
a)\(n_{HCl}=\dfrac{10,65}{36,5}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15 0,15
b)\(V_{H_2}=0,15\cdot22,4=3,36l\)
c)\(n_{CuO}=\dfrac{16}{80}=0,2mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,2 0,15 0,15
\(m_{Cu}=0,15\cdot64=9,6g\)
\(a) Mg + 2HCl \to MgCl_2 + H_2\\ b) n_{MgCl_2} = n_{Mg} = \dfrac{0,24}{24} = 0,01(mol)\\ m_{MgCl_2} = 0,01.95 = 0,95(gam)\\ c) n_{H_2} = n_{Mg} = 0,01(mol) \Rightarrow V_{H_2} = 0,01.22,4 = 0,224(lít)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{18,48}{22,4}=0,825\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}24x+27y=17,1\\x+\dfrac{3}{2}y=0.825\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,3\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,375.24}{17,1}.100=52,63\%\\ \%m_{Al}=47,37\%\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,2------------------------>0,2
=> VH2 = 0,2.24,79 = 4,958 (l)
\(nFe=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,05
\(VH_2=0,05.22,4=1,12\left(lít\right)\)
a)\(Zn++2HCl-->ZnCl2+H2\)
b)\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c)\(n_{ZnCl2}=n_{Zn}=0,2\left(mol\right)\)
\(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
d)\(n_{H2}=n_{Zn}=0,2\left(mol\right)\)
\(V_{H2}=0,2.22,4=4,48\left(l\right)\)
a. \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b. \(m_{HCl}=\dfrac{100\cdot9,6}{100}=9,6\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{9,6}{36,5}\approx0,3\left(mol\right)\)
Theo PTHH: \(n_{Mg}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0,3=0,15\left(mol\right)\)
\(\Rightarrow a=0,15\cdot24=3,6\left(g\right)\)
c.Theo PTHH: \(n_{H2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0,3=0,15\left(mol\right)\)
\(\Rightarrow V_{H2}=0,15\cdot22,4=3,36\left(l\right)\)
a) PTHH: Mg + 2HCl ->MgCl2 + H2
b) nHCl= 100:36,5=2,74(mol)
pthh: Mg + 2HCl -> MgCl2 + H2
-> nMg= 2,74÷2×1=1,37(mol)
-> mMg=1,37×24=32,88(g)
-> a=32,88(g)
c) Theo ý a) -> nH2=2,74(mol)
->VH2=2,74×22,4=61,376(l)