Đáp án:

 1. A

2. B

3. D

Giải thích các bước giải:

 1.

\(2KMnO_4\xrightarrow{t^\circ} K_2MnO_4+MnO_2+O_2\\ n_{O_2\ (pt)}=\dfrac 12\cdot n_{KMnO_4}=0,1\ \text{mol}\)

Vì hao hụt 10% nên \(n_{O_2\ (tt)}=0,1\cdot 90\%=0,09\ \text{mol}⇔ V_{O_2\ (tt)}=0,09\cdot 22,4=2,016\ \text{lít}\)

Suy ra chọn A

2.

\(2Mg+O_2\xrightarrow{t^\circ} 2MgO\)

\(\Delta_m=m_{O_2}=0,64\ \text{gam} ⇒ n_{O_2}=\dfrac{0,64}{32}=0,02\ \text{mol}\)

\(⇒n_{Mg}=2\cdot n_{O_2}=0,04\ \text{mol}⇔m=0,04\cdot 24=0,96\ \text{gam}\)

Suy ra chọn B

3.

\(2KClO_3\xrightarrow{t^\circ, MnO_2}2KCl+3O_2\)

\(n_{O_2}=\dfrac{2,688}{22,4}=0,12\ \text{mol}\)

Theo PTHH: \(n_{KClO_3}=\dfrac 23\cdot n_{O_2}=0,08\ \text{mol}\)

\(⇒m_{KClO_3}=122,5\cdot 0,08=9,8\ \text{gam}\)

Suy ra chọn D