Đốt cháy hoàn toàn 3,68 gam hỗn hợp C và S trong O2 vừa đủ

* Hướng dẫn giải

Đáp án D

Đặt số mol C, S lần lượt là x, y

$$\begin{gathered} \left\{\begin{array}{l}12\mathrm{x}+32\mathrm{y}=3,68\\ {\mathrm{m}}_{\downarrow }={\mathrm{m}}_{{\mathrm{BaCO}}_3}+{\mathrm{m}}_{{\mathrm{BaSO}}_3}=197\mathrm{x}+217\mathrm{y}=38,83\mathrm{g}\end{array}\Rightarrow \left\{\begin{array}{l}\mathrm{x}=0,12\\ \mathrm{y}=0,07\end{array}\right.\right. \\ \mathrm{Quy} \mathrm{đổi}: \left\{\begin{array}{l}\mathrm{C},\mathrm{S}\to \mathrm{X}:{\mathrm{M}}_{\mathrm{X}}=\frac{3,68}{0,12+0,07}=\frac{368}{19}\\ \mathrm{NaOH},\mathrm{KOH}\to \mathrm{ROH}\end{array}\right. \end{gathered}$$

0,19 mol X + 0,1a mol ROH ® m₁ gam muối/V₁ ml dung dịch Z

0,19 mol X + 0,16a mol ROH  ®m₂ gam muối/V₂ ml dung dịch T

6V₁ ml Z + V₂ ml T ® muối trung hòa

Þ Chứng tỏ Z chứa muối axit, T chứa kiềm dư và $6{\mathrm{n}}_{{\mathrm{HXO}}_3^{-}\left(\mathrm{Z}\right)}={\mathrm{n}}_{{\mathrm{OH}}^{-}\left(\mathrm{T}\right)}$

=> b.(0,38-0,1a)=0,16a-0,38 => a=3,5

$$\begin{gathered} {\mathrm{m}}_2-{\mathrm{m}}_1={\mathrm{m}}_{\mathrm{ROH} \left(\mathrm{T}\right)}+{\mathrm{m}}_{{\mathrm{RXO}}_3 \left(\mathrm{T}\right)}-{\mathrm{m}}_{{\mathrm{RXO}}_3 \left(\mathrm{Z}\right)}-{\mathrm{m}}_{{\mathrm{RHXO}}_3 \left(\mathrm{Z}\right)} \\ {\mathrm{m}}_2-{\mathrm{m}}_1=(\mathrm{R}+17).0,18+\left(2\mathrm{R}+\frac{1280}{19}\right).\left(0,19-0,16\right)-\left(\mathrm{R}+\frac{1299}{19}\right).0,03=8,82 \\ \Rightarrow \mathrm{R}=\frac{193}{7} \\ \Rightarrow \left\{\begin{array}{l}{\mathrm{m}}_1=\left(\frac{193}{7}+17\right).0,18+\left(2.\frac{193}{7}+\frac{1280}{19}\right).0,19=31,3 \mathrm{g}\\ {\mathrm{m}}_2=\left(2.\frac{193}{7}+\frac{1280}{19}\right).0,16+\left(\frac{193}{7}+\frac{1299}{19}\right).0,03=22,48 \mathrm{g}\end{array}\right. \\ \Rightarrow {\mathrm{m}}_1+{\mathrm{m}}_2=53,78 \mathrm{gần} \mathrm{nhất} \mathrm{với} \mathrm{giá} \mathrm{trị} 54. \end{gathered}$$