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\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b.\)
\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)
\(m_{Mg}=0.25\cdot24=6\left(g\right)\)
\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)
\(c.\)
\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6 0,2 0,3
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(C_{M\left(HCl\right)}=\dfrac{0,6}{0,6}=1\left(M\right)\)
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)
\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\\ n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\\ n_{Na}=n_{C_2H_5OH}=0,1\left(mol\right)\\ n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a=m_{Na}=0,1.23=2,3\left(g\right)\\ V=V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)
