Câu 4:
a, $Fe+ 2HCl \to FeCl_2+H_2$
b,
$n_{Fe}=\frac{5,6}{56}=0,1 mol$
$n_{HCl}=0,1 mol$
$\Rightarrow$ Dư 0,05 mol Fe.
$m_{Fe dư}=0,05.56= 2,8g$
c,
$n_{FeCl_2}=n_{H_2}=0,05 mol$
$V_{H_2}=0,05.22,4= 1,12l$
d,
$C_{M_{FeCl_2}}=\frac{0,05}{0,1}=0,5M$
Câu 5:
a, $Zn+2HCl\to ZnCl_2+H_2$
b,
$n_{Zn}=\frac{6,5}{65}=0,1 mol= n_{ZnCl_2}=n_{H_2}$
$\Rightarrow V_{H_2}=0,1.22,4=2,24l$
c,
$m_{ZnCl_2}=0,1.136= 13,6g$