\(n_{Zn}=0,2mol\\ a.2Zn+O_2-^{^{ }t^{^0}}->2ZnO\\ b.m_{ZnO}=0,2.71=14,2g\\ n_{O_2}=0,2:2=0,1mol\\ V_{O_2}=0,1.22,4=2,24L\\ c.2KClO_3-^{^{ }t^{^{ }0}}->2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.0,1=\dfrac{0,2}{3}mol\\ m_{KClO_3}=122,5\cdot\dfrac{0,2}{3}=8,166g\)
Đúng 3
Bình luận (1)