1/.
Na = 27,45/23 = 1,2
N = 16,47/14 = 1,2
O = 56,47/16 = 3,53
rezultatele le impartim la cel mai mic multiplu comun
=> Na = 1, N = 1 si O = 3
=> NaNO3, azotat de sodiu
2/.
Al + H2SO4 --> Al2(SO4)3 + H2
CuCO3 --toC--> CuO + CO2
FeCl3 + 3NaOH --> 3NaCl + Fe(OH)3↓
2Na + 2H2O --> 2NaOH + H2
3/.
ms = 160 g , c% = 40%
stim c% = mdx100/ms
=> md = msxc%/100
160x40/100 = 64 g
md g 64 g m g
FeCl3 + 3NaOH --> 3NaCl + Fe(OH)3↓
162,5 3x40 107
=> md = 162,5x64/3x40 = 86,67 g FeCl3
c% = 18,25% => ms = mdx100/c%
= 86,67x100818,25 = 475 g sol FeCl3
=> m = 64x107/3x40 = 57 g hidroxid
