Đáp án:
D
Giải thích các bước giải:
$m_{HCl}=93,65\cdot 3,9\%=3,65235\ (gam)$
$⇒n_{HCl}=\dfrac{3,65}{36,5}=0,1\ (mol)$
$n_{H_2O}=\dfrac{93,65-3,65}{18}=5\ (mol)$
PTHH:
$2Na+2HCl\to 2NaCl+H_2$
$Ba+2HCl\to BaCl_2+H_2$
$2Na+2H_2O\to 2NaOH+H_2$
$Ba+2H_2O\to Ba(OH)_2+H_2$
Theo PTHH, ta có: $n_{H_2}=\dfrac 12n_{HCl}+\dfrac 12n_{H_2O}=2,55\ (mol)$
$⇒V=2,55.22,4=57,12\ (l)$
$⇒$ Chọn D