Em tham khảo, anh có làm rồi nè
$n_{SO_2} = \dfrac{16}{32} = 0,5(mol)$
$n_{Ba(OH)_2} = \dfrac{300.17,1\%}{171} = 0,3(mol)$
\(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O\)
a a a (mol)
\(2SO_2+Ba\left(OH\right)_2\rightarrow Ba\left(HSO_3\right)_2\)
2b b b (mol)
Ta có:
$a + b = 0,3$
$a + 2b = 0,5$
Suy ra : a = 0,1 ; b = 0,2
$m_{dd} = 16 + 300 - 0,1.217 = 294,3(gam)$
$C\%_{Ba(HSO_3)_2} = \dfrac{0,2.299}{294,3}.100\% = 20,32\%$
\(n_{SO_2}=\dfrac{16}{64}=0,25\left(mol\right)\)
\(m_{Ba\left(OH\right)_2}=300.17,1\%=51,3\left(g\right)\Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{51,3}{171}=0,3\left(mol\right)\)
PTHH: SO2 + Ba(OH)2 → BaSO3 + H2O
Mol: 0,25 0,25 0,25
Ta có: \(\dfrac{0,25}{1}< \dfrac{0,3}{1}\) ⇒ SO2 pứ hết, Ba(OH)2 dư
mdd sau pứ = 16+300 = 316 (g)
\(C\%_{ddBaSO_3}=\dfrac{0,25.217.100\%}{316}=17,17\%\)
\(C\%_{ddBa\left(OH\right)_2dư}=\dfrac{\left(0,3-0,25\right).171.100\%}{316}=2,7\%17,17\%\)
