80 g roztworu NaOH o stężeniu 20% reaguje z kwasem siarkowym VI . Ile g soli powstaje ?

M(NaOH)=23g+16g+1g=40g M(Na₂SO₄)=2*23g+32g+4*16g=142g   mr=80g Cp=20% ms=?   Cp=ms/mr * 100%    /*mr Cp*mr=ms*100%    /:100% ms=Cp*mr/100%   ms=20%*80g/100% ms=16g   2NaOH+H₂SO₄-->Na₂SO₄+2H₂O 16g-------------------xg 80g-------------------142g   xg=16g*142g/80g xg=28,4g