a) Al= (27/132)*100%= 20.5% Cl3= (35*3)/132*100%=79.5% b) Na= 23/119*100%=19% S=32/119*100%=27% O4=64/119*100%=54% c) Pb=207/333*100%=62% N2= 28/333*100%=8% O6=96/333*100%=30%
a)AlCl3 m AlCl3 = 27u + 3*35u = 27u + 105u = 132 m Al = 27u 132u - 100% 27u - x x = 27u * 100%/ 132 x = 20,45% mCl3 = 105u 132u - 100% 105u - x x = 105*100%/132u x = 79,54% b)NaSO4 mNaSO4 = 23u + 32u + 4*16u = 119u mNa = 23u 119u - 100% 23u - x x = 23u*100%/119u x = 19,32% mSO4 = 96u 119u - 100% 96u - x x = 96u*100%/119u x = 80,67% c)Pb(NO3)2 mPb(NO3)2 = 207u + (14u + 3*16u)*2 = 207u + 124u = 331 mPb = 207u 331u - 100% 207u - x x = 207u*100%/331u x = 62,53% m(No3)2 = 124u 331u - 100% 124u - x x = 124u*100%/331u x = 38,47%
