\(n_{Zn}=\dfrac{15,6}{65}=0,24\left(mol\right)\)
\(Zn+2HCl\underrightarrow{ }ZnCl_2+H_2\)
mol 1: 2 : 1 : 1
mol 0,24:0,48:0,24:0,24
\(Zn+H_2SO_4\underrightarrow{ }ZnSO_4+H_2\)
mol 1 1 1 1
mol 0,24: 0,24:0,24:0,24
\(m_{ct\left(HCl\right)}=0,24.36,5=8,76\left(g\right)\)
\(m_{ct\left(H_2SO_4\right)}=0,24.98=23,52\left(g\right)\)
\(m_{dd\left(HCl\right)}=\dfrac{8,76}{14,6\%}.100\%=60\left(g\right)\)
\(m_{dd\left(H_2SO_4\right)}=\dfrac{23,52}{19,6\text{%}}.100\%=120\left(g\right)\)
mình lại xíu
\(m_{ct\left(HCl\right)}=0,48.36,5=17,52\left(g\right)\)
\(m_{dd\left(HCl\right)}=\dfrac{17,52}{14,6\%}.100\%=120\left(g\right)\)
\(n_{Zn}=\dfrac{15,6}{65}=0,24\left(mol\right)\)
m(g) dung dịch hỗn hợp \(|^{HCl 14,6\%}_{H_2SO_4 19,6\%}\)
\(n_{HCl}=\dfrac{14,6m}{100.36,5}=0,004m\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{19,6m}{100.98}=0,002m\left(mol\right)\)
PTHH:
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2\(\uparrow\) (1)
(mol) 0,002m....0,004m
Zn + H2SO4 \(\rightarrow\) ZnSO4 + H2\(\uparrow\) (2)
(mol) 0,002m...0,002m
\(\left(1\right),\left(2\right)\rightarrow n_{Zn}=0,002m+0,002m=0,24\)
\(\rightarrow m=60\left(g\right).\)
