$n_{H_2SO_4}=0,07$
$n_{BaSO_4}=0,04=n_{Ba(OH)_2}$
$Ba(OH)_2 +H_2SO_4→BaSO_4+2H_2O$
$0,04$ $0,04$ $0,04$
$2NaOH +H_2SO_4→Na_2SO_4+2H_2O$
$0,06$ $0,03$
$Cm_{NaOH}=\frac{0,06}{0,05}=1,2M$
$Cm_{Ba(OH)_2}=\frac{0,04}{0,05}=0,8M$
$n_{Al}=0,04$
$Al +2OHAlO_2 +H_2$
$0,04$ $0,08$
$\text{Gọi thể tích B là V}$
$\text{Ta có:}$ $0,02·1,2+V·0,8·2=0,08→V=0,035l=35ml$