Zn + 2HCl → ZnCl2 + H2
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(m_{HCl}=65\times20\%=13\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{13}{36,5}=\dfrac{26}{73}\left(mol\right)\)
Theo PT: \(n_{Zn}=\dfrac{1}{2}n_{HCl}\)
theo bài: \(n_{Zn}=\dfrac{73}{130}n_{HCl}\)
Vì \(\dfrac{73}{130}>\dfrac{1}{2}\) ⇒ Zn dư
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times\dfrac{26}{73}=\dfrac{13}{73}\left(mol\right)\)
\(\Rightarrow m_{H_2}=\dfrac{13}{73}\times2=0,36\left(g\right)\)
\(\Rightarrow m_{ddZnCl_2}=13+65-0,26=77,64\left(g\right)\)
theo PT: \(n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times\dfrac{26}{73}=\dfrac{13}{73}\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=\dfrac{13}{73}\times136=24,22\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{24,22}{77,64}\times100\%=31,2\%\)
Zn + 2HCl → ZnCl2 + H2
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(m_{HCl}=65\times20\%=13\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{13}{36,5}=\dfrac{26}{73}\left(mol\right)\)
Theo PT: \(n_{Zn}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Zn}=\dfrac{73}{130}n_{HCl}\)
Vì \(\dfrac{73}{130}>\dfrac{1}{2}\) ⇒ Zn dư
Theo PT: \(n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times\dfrac{26}{73}=\dfrac{13}{73}\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=\dfrac{13}{73}\times136=24,22\left(g\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times\dfrac{26}{73}=\dfrac{13}{73}\left(mol\right)\)
\(\Rightarrow m_{H_2}=\dfrac{13}{73}\times2=0,36\left(g\right)\)
\(m_{dd}saupư=13+65-0,36=77,64\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{24,22}{77,64}\times100\%=0,31\%\)
nZn = \(\dfrac{13}{65}\) = 0,2 mol
mHCl = \(\dfrac{65.20}{100}\) = 13 g => nHCl = \(\dfrac{13}{36,5}\) = \(\dfrac{26}{73}\)mol
a) Zn +2 HCl -> ZnCl2 + H2
0,2(dư);\(\dfrac{26}{73}\)(hết)->\(\dfrac{13}{73}\)
=>mZnCl2 = \(\dfrac{13}{73}.138\)= 24,5g
=>C%ZnCl2 = \(\dfrac{24,5}{13+65}.100\text{%}\) \(\approx\) 331,4 %
