Calculez masa de acid din solutie, apoi, din ecuatie,cantitatile de sare si apa
M,H2SO4=98g/mol, M,MgO= 40g/mol, M,MgSO4=120g/mol, M,H20=18g/mol
c= mdx100/ms---> md= 98x200/100g=2x98g H2SO4...>.2mol
1mol.........1mol..........1mol.........1mol
MgO + H2SO4----> MgSO4 + H2O
X=2.mol......2..............2.................2
m,MgSO4= 2molx120g/mol=240g
m,H2O=2molx18g/mol=36g
calculez conc. sarii, stiind md=240g,
m,apa= m,apa ramasa din sol. de acid + m,apa rezultata din reactie
(200-196)g +36g= 40g
ms= md+m,apa= 240g+40g=280g
c= 240x100/280=...........................calculeaza !!
1mol molec...contine....6,023x10²³molec
2mol...............................N= 12,046x10²³molec ,rezultat valabil si pentru MgSO4, cat si pentru H2O
