fiind vb. de un cristalohidrat consideram ca 57,2 g = ms
=> md = masa de sare anhidra
=> M.molec. Na2CO3 x 10H2O = 106+180 = 286 g/mol
=> daca la 286 g cristal. ....................... 106 g sare
la 57,2 g ......................................a = 21,2 g sare Na2CO3
21,2 g......... md......................................... V L
Na2CO3 + 2HCl => CaCl2 + H2O + CO2
106.................2x36,5..................................22,4
=> md = 21,2x2x36,5/106 = 14,6 g HCl
stim ms = 350 g si c% = mdx100/ms
=> c = 14,6x100/350 = 4,17%
=> V = 21,2x22,4/106 = 4,48 L CO2
