Đáp án:

 

Giải thích các bước giải:

$HCHO + O_{2}\xrightarrow{t^o}CO_{2} + H_{2}O$ (1)

$C_{2}H_{5}CHO + 4O_{2}\xrightarrow{t^o}3CO_{2} + 3H_{2}O$ (2)

$CO_{2} + Ba(OH)_{2}→BaCO_{3} + H_{2}O$ (3)

Ta có: $n_{HCHO}=\dfrac{0,6}{30}= 0,02 (mol)$

$n_{C_{2}H_{5}CHO}=\dfrac{1,74}{58}= 0,03 (mol)$

Theo PT (1): $n_{CO_{2}}=n_{HCHO}= 0,02 (mol)$

Theo PT (2): $n_{CO_{2}}=3n_{C_{2}H_{5}CHO}=3 × 0,03= 0,09 (mol)$

⇒ ∑$n_{CO_{2}}=0,02 + 0,09= 0,11 (mol)$

⇒ $m_{CO_{2}}=0,11 × 44= 4,84 (g)$

Theo PT (3): $n_{Ba(OH)_{2}}=n_{CO_{2}}= 0,11 (mol)$

⇒ $m_{Ba(OH)_{2}}=0,11 × 197= 21,67 (g)$

$m_{\text{dd sau pứ}}=4,84 + m_{\text{dd Ba(OH)2}} - 21,67=m_{\text{dd Ba(OH)2}} - 16,83$

⇒ dung dịch $Ba(OH)_{2}$ giảm $16,83g$

Em tham khảo nha :

\(\begin{array}{l}
HCHO + {O_2} \to C{O_2} + {H_2}O\\
{C_2}{H_5}CHO + 4{O_2} \to 3C{O_2} + 3{H_2}O\\
{n_{HCHO}} = \dfrac{{0,6}}{{30}} = 0,02mol\\
{n_{{C_2}{H_5}CHO}} = \dfrac{{1,74}}{{58}} = 0,03mol\\
{n_{C{O_2}}} = {n_{HCHO}} + 3{n_{{C_2}{H_5}CHO}} = 0,11mol\\
C{O_2} + Ba{(OH)_2} \to BaC{O_3} + {H_2}O\\
{n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,11mol\\
{m_{{\rm{dd}}spu}} = 0,11 \times 44 + {m_{{\rm{dd}}Ba{{(OH)}_2}}} - 0,11 \times 197 = {m_{{\rm{dd}}Ba{{(OH)}_2}}} - 16,83
\end{array}\)

Vậy dung dịch $Ba(OH)_2$ giảm 16,83g