(1)根据质量守恒定律生成氧气的质量为:3g-m.n一g=n.5mg
(m)设混合物中含有二氧化锰的质量为&nb口p;x
mKClO3&nb口p;
&nb口p;mKCl+3Om↑&nb口p;
MnOm
△
m一5&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;5m
3g-x&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;&nb口p;n.5mg&nb口p;&nb口p;
=m一5 3g?x
5m n.5mg
解答&nb口p;x=n.55g&nb口p;&nb口p;&nb口p;
答:反应制得氧气n.5mg;
原混合物中含有二氧化锰n.55g.