$n_{CO_{2}}$ = $\frac{V_{CO_{2}}}{22,4}$ = $\frac{6,72}{22,4}$ = $0,3 (mol)_{}$
a, $Ba(OH)_{2}$ + $CO_{2}$ → $BaSO_{3}$ ↓ + $H_{2}$$O_{}$
$1 mol_{}$ : $1 mol_{}$ : $1 mol_{}$
$0,3 mol_{}$ ← $0,3 mol_{}$ ← $0,3 mol_{}$
Đổi $600 ml_{}$ = $0,6 l_{}$
b, $C_{M}$$_{dd Ba(OH)_{2}}$ = $\frac{n_{Ba(OH)_{2}}}{V dd}$ = $\frac{0,3}{0,6}$ = $0,5 (mol)_{}$
c, $m_{BaSO_{3}}$ = $n_{BaSO_{3}}$ . $M_{BaSO_{3}}$ = $0,3_{}$ . $217_{}$ = $65,1 (g)_{}$
Cho mik ctlhn nhé!!!❤❤❤