ms = 350g , c% = 70%
stim ca c% = mdx100/ms
=> md = msxc%/100 = 350x70/100 = 245g H2SO4
M.H2SO4 = 2x1+32+4x16 = 98 g/mol
M.BaCl2 = 137+2x35,5 = 208 g/mol
245g m g
H2SO4 + BaCl2 --> BaSO4 + 2HCl
98 208
=> m = 245x208/98 = 520 g BaCl2 pur consumat teoretic la 100% ... dar noi aven randament de 90% deci practic cat s-a consumat
100% .......................... 520g BaCl2 (Ct)
90% ............................ Cp = 468 g BaCl2 pur consumat practic
75% ...................... 568g BaCl2 pur (m.pura)
100% ..................... m.impura = 624 g BaCl2
