Đáp án:
a) PTHH $Zn+2HCl→ZnCl_2+H_2$
b) $m_{ddHCl}=100(g)$
c) $m_{ZnCl_2}=13,6(g)$
Giải thích các bước giải:
a) $PTHH:$ $Zn+2HCl→ZnCl_2+H_2$
b) $n_{Zn}=\dfrac{6,5}{65}=0,1(mol)$
$n_{HCl}=2n_{Zn}=2×0,1=0,2(mol)$
$m_{HCl}=0,2×36,5=7,3(g)$
$\dfrac{7,3}{m_{ddHCl}}×100=7,3$
⇒ $m_{ddHCl}=100(g)$
c) $n_{ZnCl_2}=n_{Zn}=0,1(mol)$
$m_{ZnCl_2}=0,1×136=13,6(g)$