\(a) n_{KMnO_4} = \dfrac{31,6}{158} = 0,2(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2(1)\\ n_{O_2} = \dfrac{1}{2}n_{KMnO_4} = 0,1(mol) \Rightarrow m_{O_2} = 0,1.32 = 3,2(gam)\\ b) Coi\ m_{KMnO_4} = m_{KNO_3} = 100(gam)\\ 2KNO_3 \xrightarrow{t^o} 2KNO_2 + O_2(2)\\ n_{O_2(1)} = \dfrac{1}{2}n_{KMnO_4} = \dfrac{1}{2}.\dfrac{100}{158} = \dfrac{25}{79}(mol)\\ n_{O_2(2)} = \dfrac{1}{2}n_{KNO_3} = \dfrac{1}{2}.\dfrac{100}{101} = \dfrac{50}{101}(mol)\\ Vì\ n_{O_2(1)} < n_{O_2(2)}\ \text{nên đốt}\ KNO_3\ \text{thu được nhiều khí hơn}\)
Đúng 3
Bình luận (1)