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a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)
c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)
Ta có: m dd sau pư = 1,2 + 36,5 - 0,05.2 = 37,6 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)
a. 2Al + 6HCl -> 2AlCl3 + 3H2
b. nAl = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)
\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)
\(n_{H_2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
_________\(0,2\)____\(0,6\)_____\(0,2\)______\(0,3\left(mol\right)\)
a) \(m_{Al}=0,2.27=5,4\left(g\right)\)
b) \(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(C\%_{HCl}=\frac{21,9}{200}.100\%=10,95\%\)
c) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(C\%_{AlCl_3}=\frac{26,7}{200}.100\%=13,35\%\)
d) \(V_{AlCl_3}=\frac{26,7}{1,1}=\frac{267}{11}\left(l\right)\)
\(C_{M_{AlCl_3}}=\frac{0,2}{\frac{267}{11}}=0,007\left(M\right)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{Zn}=n_{H_2}=0,15\left(mol\right);n_{HCl}=2.0,15=0,3\left(mol\right)\\ a,m=m_{Zn}=0,15.65=9,75\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,3}{0,15}=0,2\left(l\right)\\ c,m_{ZnCl_2}=0,15.136=20,4\left(g\right)\)
\(V_{H_2}\)= \(\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)PTHH: 2Al+6HCL→2AlCl3+3H2a)Theo pt: \(n_{Al_{ }}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2mol\)⇒ mAl=n.M=0,2.27=5,4gb)Theo pt:\(n_{HCl}=\dfrac{6}{2}n_{H_2}=\dfrac{6}{2}.0,3=0,9mol\)⇒mHCl=n.M=0,9.36.5=32,85g
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(\Leftrightarrow n_{HCl}=0.3\left(mol\right)\)
\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Vì:\dfrac{0,2}{2}>\dfrac{0,15}{3}\Rightarrow Aldư\\ \Rightarrow n_{HCl}=2.n_{H_2}=2.0,15=0,3\left(mol\right)\\ \Rightarrow m_{HCl}=0,3.36,5=10,95\left(g\right)\)
\(a.n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,3 0,1 0,15
\(m_{Al}=0,1.27=2,7g\\ b.C_{M_{HCl}}=\dfrac{0,3}{0,3}=1M\\ c.C_{\%HCl}=\dfrac{0,3.36,5}{300.1,2}\cdot100=3,04\%\\ d)m_{dd}=2,7+300.1,2-0,15.2=362,4g\\ C_{\%AlCl_3}=\dfrac{0,1.133,5}{362,4}\cdot100=3,68\%\)