Đáp án:
a) mFe=1,68 gam; V O2=0,448 lít
b) mKMnO4=6,32 gam
Giải thích các bước giải:
a) \(3Fe + 2{O_2}\xrightarrow{{}}F{e_3}{O_4}\)
Ta có: \({n_{F{e_3}{O_4}}} = \frac{{2,32}}{{56.3 + 16.4}} = 0,01{\text{ mol}} \to {{\text{n}}_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,03{\text{ mol; }}{{\text{n}}_{{O_2}}} = 2{n_{F{e_3}{O_4}}} = 0,02{\text{ mol}}\)
\(\to {m_{Fe}} = 0,03.56 = 1,68{\text{ gam; }}{{\text{V}}_{{O_2}}} = 0,02.22,4 = 0,448{\text{ lít}}\)
b) \(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\(\to {n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,02.2 = 0,04{\text{ mol}} \to {{\text{m}}_{KMn{O_4}}} = 0,04.(39 + 55 + 16.4) = 6,32{\text{ gam}}\)
