$\#duaconngoaihanhlang$

$PTHH$ $:$

$Ba(OH)_2$ $+$ $H_2SO_4$ $→$ $BaSO_4↓$ $+$ $2H_2O$ $(1)$

$BaCO_3$ $+$ $H_2SO_4$ $→$ $BaSO_4↓$ $+$ $CO_2↑$ $+$ $H_2O$ $(2)$

$⇒$ $Kết$ $tủa$ $là$ $BaSO_4$

$n_{H_2SO_4}$ $=$ $0,25$ $×$ $1$ $=$ $0,25$ $(mol)$

$Theo$ $(1)$ $,$ $(2)$ $:$ $n_{BaSO_4}$ $=$ $n_{H_2SO_4}$ $=$ $0,25$ $(mol)$

$⇒$ $m_{BaSO_4}$ $=$ $0,25$ $×$ $233$ $=$ $58,25$ $(g)$

$Theo$ $(1)$ $:$ $n_{H_2SO_4}$ $_{(1)}$ $=$ $n_{Ba(OH)_2}$ $=$ $0,1$ $(mol)$

$⇒$ $n_{H_2SO_4}$ $_{(2)}$ $=$ $0,25$ $-$ $0,1$ $=$ $0,15$ $(mol)$

$Theo$ $(2)$ $:$ $n_{CO_2}$ $=$ $n_{H_2SO_4}$ $_{(2)}$ $=$ $0,15$ $(mol)$

$⇒$ $V_{CO_2}$ $=$ $0,15$ $×$ $22,4$ $=$ $3,36$ $(g)$

\(\begin{array}{l}
250ml=0,25l\\
n_{H_2SO_4}=0,25.1=0,25(mol)\\
Ba(OH)_2+H_2SO_4\to BaSO_4+2H_2O\\
BaCO_3+H_2SO_4\to BaSO_4+CO_2+H_2O\\
Theo\,PT:\,\sum n_{BaSO_4}=\sum n_{H_2SO_4}=0,25(mol)\\
\to m_{ket\,tua}=m_{BaSO_4}=0,25.233=58,25(g)\\
Theo\,PT:\,n_{CO_2}=\sum n_{H_2SO_4}-n_{Ba(OH)_2}=0,15(mol)\\
\to V_{CO_2}=0,15.22,4=3,36(l)
\end{array}\)