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1
\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
a
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,3-->0,6---->0,3------->0,3
b
\(C\%_{dd.HCl}=\dfrac{0,6.36,5.100\%}{400}=5,475\%\)
c
\(m_{dd}=16,8+400-0,3.2=416,2\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{0,3.127.100\%}{416,2}=9,15\%\)
2
\(n_{HCl}=\dfrac{200.7,3\%}{100\%}:36,5=0,4\left(mol\right)\)
a
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2<--0,4------>0,2------>0,2
b
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c
\(x=m_{Mg}=0,2.24=4,8\left(g\right)\)
d
\(m_{dd}=4,8+200-0,2.2=204,4\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{0,2.95.100\%}{204,4}=9,3\%\)
Câu 15 :
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,4----->1,2------->0,4------>0,6
\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{43.8.100\%}{25\%}=175,2\left(g\right)\)
\(m_{ddspu}=10,8+175,2-0,6.2=184,8\left(g\right)\)
\(C\%_{AlCl3}=\dfrac{0,4.133,5}{184,8}.100\%=28,9\%\)
a) $CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
Theo PTHH :
$n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$
b) $n_{HCl} = 2n_{CaCO_3} = 0,2(mol)$
$C_{M_{HCl}} = \dfrac{0,2}{0,25} = 0,8M$
c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,1(mol)$
$m_{CaCO_3} = 0,1.100 = 10(gam)$
a,\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,1 0,2 0,1 0,1
b,\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
c,\(m_{ddHCl}=\dfrac{0,2.36,5.100}{3,65}=200\left(g\right)\)
d,\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
e,mdd sau pứ = 6,5+200-0,1.2 = 206,3 (g)
\(C\%_{ddZnCl_2}=\dfrac{13,6.100\%}{206,3}=6,59\%\)
Câu 3 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{10,95}=133,3\left(g\right)\)
c) \(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
d) \(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{MgCl2}=0,2.95=19\left(g\right)\)
\(m_{ddspu}=4,8+133,3-\left(0,2.2\right)=137,7\left(g\right)\)
\(C_{MgCl2}=\dfrac{19.100}{137,7}=13,8\)0/0
Chúc bạn học tốt
a)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
b)
\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)
Theo PT:\(n_{HCl}=2n_{Fe}=0,2mol\)
\(\Rightarrow C_MHCl=\dfrac{0,2}{0,2}=1M\)
c)
Theo PT:\(n_{H_2}=n_{Fe}=0,1mol\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24l\)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
_____0,1--->0,2
=> \(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)\)