1) najpierw obliczamy mase molową całej soli a) AlCl3=133,5g/mol 133,5-100% 27-x% x=20,2% wiec chlor 100-20,2=79,8% b)Na2So4= 142 g/ mol 142-100% 46-x x= 32,4% sód 142-100% 32-x x=22,5% siarka 100-22,5-32,4=45,1% tlen C) Pb(NO3)2= 331g/mol 331-100% 207-x x=62,5% ołów 331-100% 28-x x=8,45% azot 100-62,5-8,45=29,05% tlen
a)AlCl₃ mAl=27u mCl=35,5u mAlCl3=133,5u 133,5g---100% 27g-----x% x=20,2% Al 133,5g---100% 106,5g-----x% x=79,8% Cl b)Na₂SO₄ mNa=23u mS=32u mO=16u mNa₂SO₄=142u 142g----100% 46g-----x% x=32,4% Na 142g----100% 64g-----x% x=45% O 142g----100% 32g-----x% x=22,6% S c)Pb(NO₃)₂ mPb=207u mN=14u mO=16u mPb(NO₃)₂=331u 331g-----100% 207-------x% x=62,5% Pb 331g-----100% 28-------x% x=8,5% N 331g-----100% 96-------x% x=29% O
a) mAlCL3 = 1*27 + 3*35,5 = 27+ 106,5= 133,5u 133,5 - 100% 27 - x x=2700/133,5 = 20,2 % - Al 100 - 20,2 = 79,8% - Cl b)mNa2SO4 = 2*23+ 1*32 + 4*16 = 36+32+64= 132u 132 - 100% 64 - x x=6400/132 = 48,5% - O 132 - 100% 32 - x x=3200/132 = 24,3 % - S 100 - (24,3+48,5) = 27,2% - Na c) mPB(NO3)2= 1*207 + 2*14 + 6*16 = 207+28+96= 331u 331 - 100% 207 - x x= 20700/331 = 62,5% - Pb 331 - 100% 96 - x x= 9600/331= 29% - O 100 - (62,5 +29) = 8,5 % - N
