Na2CO3
a) nNa : nC : nO = 2 : 1 : 3
b) mNa: mC : mO = 2·23 : 12 : 3·16 = 46 : 12 : 48 = 23 : 6 : 24
c) M = 2·23 + 1·12 + 3·16 = 106g/mol
%Na = 46·100/106 = 43,39% %C = 12·100/106 = 11,32%
% O = 48·100/106 = 45,28%
d) n = 18,069·10²²/(6,023·10²³) = 0,3 moli
m = nxM = 0,3x106 = 31,8g
ii ok?