Bạn tham khảo nhé
a) Ta có: Fe=56 => MFe = 56 (g)
Fe2O3 = 2*56 + 3*16 = 160
=>MFe2O3 = 160 (g)
=> %mFe = (2*56)/ 160 * 100 % = 70 (%)
%mO = 100% - 70% = 30 (%)
b) Ta có: MCa = 40 (g)
MN = 14 (g)
MCa(NO)2 = 40 +(14 + 16)*2 = 100 (g)
=> %mCa = (40/ 100) *100% = 40 (%) ;
%mN = (14*2)/100 * 100 % = 28(%) ;
%mO = (16*2)/100 * 100 % = 32 (%).
c)Ta có: MNa = 22 (g)
MC = 12 (g)
MNa2CO3 = 106 (g)
=> %Na = (23*2)/106 * 100% = 43, 39%
%C = 12/106 * 100% = 11, 32%
%O = 100% − (43, 39% + 11, 32%) = 45, 29%