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a) \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,05->0,1----->0,05---->0,05
`=> V_{ddHCl} = (0,1)/2 = 0,05 (l)`
b) `V_{H_2} = 0,05.22,4 = 1,12 (l)`
c) `C_{M(FeCl_2)} = (0,05)/(0,05) = 1M`
nFe = 5.6/56 = 0.1 (mol)
nHCl = 0.2*2 = 0.4 (mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
LTL : 0.1/1 < 0.4/2 => HCl dư
mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g)
VH2 = 0.2*22.4 = 4.48 (l)
CM FeCl2 = 0.1/0.2 = 0.5(M)
CM HCl dư = 0.2 / 0.2 = 1(M)
a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,2}{2}=0,1\left(l\right)\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,1 0,2 0,1
\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(b,V=\dfrac{n}{C_M}=\dfrac{0,2}{2}=0,1\left(l\right)\)
a,\(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,5 1 0,5 0,5
\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)\)
b,\(V_{H_2}=0,5.22,4=11,2\left(l\right)\)
c,\(C_{M_{ddFeCl_2}}=\dfrac{0,5}{0,5}=1M\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ a.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ 0,2.........0,6........0,2.........0,3\left(mol\right)\\ b.C\%_{ddHCl}=\dfrac{0,6.36,5}{200}.100=10,95\%\\ \Rightarrow a=10,95\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c.m_{ddsau}=5,4+200-0,3.2=204,8\left(g\right)\\ C\%_{ddAlCl_3}=\dfrac{133,5.0,2}{204,8}.100\approx13,037\%\)
a)
\(Zn + 2HCl \to ZnCl_2 + H_2\\ n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{19,5}{65} =0,3(mol)\\ V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{ZnCl_2} = 0,3.136 = 40,8(gam)\\ c) n_{HCl} = 2n_{Zn} = 0,6(mol) \Rightarrow V_{dd\ HCl} = \dfrac{0,6}{2} = 0,3(lít)\\ d) 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{O_2} = \dfrac{1}{2} V_{H_2} = 3,36(lít)\\ V_{không\ khí} = \dfrac{V_{O_2}}{20\%}= \dfrac{3,36}{20\%} = 16,8(lít)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15
\(a,V_{H_2}=0,15.22,4=3,36\left(l\right)\)
\(b,C_{M_{HCl}}=\dfrac{n}{V}=\dfrac{0,3}{0,1}=3M\)
\(n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{HCl} =2n_{Fe} = 0,05.2 = 0,1(mol)\\ V_{dd\ HCl} = \dfrac{0,1}{2} = 0,05(lít)\)
Bài 1:
Ta có: \(n_{HCl}=1,5\cdot0,08=0,12\left(mol\right)\) \(\Rightarrow V_{HCl\left(2M\right)}=\dfrac{0,12}{2}=0,06\left(l\right)=60\left(ml\right)\)
Bài 2:
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{HCl}=2,5\cdot2=5\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{5}{3}\left(mol\right)\\n_{H_2}=2,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=\dfrac{5}{3}\cdot27=45\left(g\right)\\V_{H_2}=2,5\cdot22,4=56\left(l\right)\end{matrix}\right.\)