Đốt cháy m gam este etyl axetat, hấp thụ hết sản phẩm vào 200 gam

* Hướng dẫn giải

Đáp án D

$\begin{array}{l}+ \left\{\begin{array}{l}\underset{x mol}{\underbrace{C{H}_{3}C\mathrm{OO}{C}_2{H}_5}}\stackrel{O_2,t^0}{\to }\underset{4x mol}{\underbrace{4C{O}_2}}+\underset{4x mol}{\underbrace{4H_{2}O}}\\ \underset{y}{\underbrace{n_{BaC{O}_3}}}+\underset{?=(0,08-y)}{\underbrace{n_{Ba{\left(HC{O}_3\right)}_2}}}=\underset{0,08}{\underbrace{n_{Ba{\left(OH\right)}_2}}}\end{array}\right.\\ + \left\{\begin{array}{l}{n}_{C{O}_2}=\underset{y}{\underbrace{n_{BaC{O}_3}}}+2\underset{(0,08-y)}{\underbrace{n_{Ba{\left(HC{O}_3\right)}_2}}}=4x\\ m_{\mathrm{dd} spư}=\underset{200}{\underbrace{m_{dd Ba{\left(OH\right)}_2}}}+\underset{248x}{\underbrace{m_{(C{O}_2, H_{2}O)}}}-\underset{197y}{\underbrace{m_{BaC{O}_3}}}=194,38\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,025; y=0,06\\ m_{C{H}_{3}C\mathrm{OO}{C}_2{H}_5}=2,2 gam\end{array}\right.\end{array}$