n moli NaxHyPO4 M1 = 23x + y + 95 x + y = 3
n moli Na2SO4 M2 =142g/mol mSO4 = 96n
96n·100/[n(M1 + 142)] = 33,8 M1 + 142 = 284 M1 = 142g/mol
23x + y = 47
x+y = 3
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22x = 44 x = 2 y = 1 Na2HPO4
amestec : m = 142n +142n = 284n mNa= 23(2n + 2n) = 92n
%Na = 92n··100/284n = 32,39%
%Na2HPO4 = 142n·100/284n = 50% (C)
%H = 100n/284n = 0,35 %
% Na2SO4 = 142n··100/284n = 50%