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a, \(n_{HCl}=0,15.1=0,15\left(mol_{ }\right)\)
PTHH: CuO + 2HCl → CuCl2 + H2O
Mol: 0,075 0,15 0,075
b, \(C_{M_{ddCuCl_2}}=\dfrac{0,075}{0,15}=0,5M\)
\(PTHH:4Al+6HCl\rightarrow2Al_2Cl_3+3H_2\uparrow\)
\(n_{Al}=\frac{3,78}{27}=0,14\left(mol\right)\)
\(\Rightarrow n_{H_2}=\frac{3}{4}n_{Al}=0,105\left(mol\right)\)
\(V_{H_2}=0,105.22,4=2,352\left(l\right)\)
\(n_{HCl}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,14=0,21\left(mol\right)\)
\(C_{M_{ddHCl}}=\frac{0,21}{0,2}=1,05\left(M\right)\)
\(n_{Al_2Cl_3}=\frac{1}{2}n_{Al}=\frac{1}{2}.0,14=0,07\left(mol\right)\)
\(m_{Al_2Cl_3}=0,07.160,5=11,235\left(g\right)\)
nHCl=0,15 . 2=0,3(mol)
PTHH: CuO + 2 HCl -> CuCl2 + H2O
x___________2x____x(mol)
ZnO +2 HCl -> ZnCl2 + H2O
y______2y_____y(mol)
Ta có hpt: \(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
Vddsau=VddHCl=0,15(l)
=> CMddCuO= 0,05/0,15=1/3(M)
CMddZnO=0,1/0,15=2/3(M)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PT: \(n_{KOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)
200ml = 0,2l
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,2 0,2
\(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(C_{M_{FeCl2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
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