n H2 = 4,48/22,4 = 0,2(mol)
Gọi n là hóa trị của A
$2A + 2nHCl \to 2ACl_n + nH_2$
Theo PTHH :
n A = 2/n nH2 = 0,4/n(mol)
=> A . 0,4/n = 13
<=> A = 65n/2
Với n = 2 thì A = 65(Zn)
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH :
n ZnCl2 = n H2 = 0,2(mol)
n HCl = 2n H2 = 0,4(mol)
=> mdd HCl = 0,4.36,5/7,3% = 200(gam)
Sau phản ứng :
mdd = m Zn + mdd HCl - m H2 = 13 + 200 - 0,2.2 = 212,6(gam)
Suy ra :
C% ZnCl2 = 0,2.136/212,6 .100% = 12,79%
\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(2A+nHCl\rightarrow2ACl_n+nH_2\)
\(\dfrac{0.4}{n}....0.4......\dfrac{0.4}{n}......0.2\)
\(M_A=\dfrac{13}{\dfrac{0.4}{n}}=32.5n\)
\(n=2\Rightarrow A=65\)
\(A:Zn\)
\(m_{dd_{HCl}}=\dfrac{0.4\cdot36.5\cdot100}{7.3}=200\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=13+200-0.2\cdot2=212.6\left(g\right)\)
\(C\%_{ZnCl_2}=\dfrac{0.2\cdot95}{212.6}\cdot100\%=8.94\%\)
\(PTHH:2A+2nHCl\rightarrow2ACl_n+nH_2\)
pt:______\(2A\left(g\right)\)________________\(22,4n\left(l\right)\)
pứ:_____\(13\left(g\right)\)_________________\(4,48\left(l\right)\)
\(\Rightarrow\dfrac{2A}{13}=\dfrac{22,4n}{4,48}\Leftrightarrow A=32,5n\Rightarrow\left\{{}\begin{matrix}A=65\left(Zn\right)\\n=2\end{matrix}\right.\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
(mol)___0,2____0,4_____0,2_____0,2__
\(C\%_{ddZnCl_2}=\dfrac{0,2.136}{13-0,2.2+\dfrac{0,4.36,5.100}{7,3}}.100=12,79\left(\%\right)\)