1.
a.
100% otel ............................ (100-1,5=)98,5% Fe
1 kg otel ........................................ a = 0,985 kg Fe
100% fonta ........................... 95% Fe
2,5 kg .......................................... b = 2,375 kg Fe
=> m.Fe.amaestec.aliaje = a+b = 0,985+2,375 = 3,36 kg Fe
b.
m.amestc.aliaje = 1+2,5 = 3,5 kg aliaje
%Fe = 3,36x100/3,5 = 96% Fe
%C = 100-96 = 4% C
2.
consideram 5g NaHCO3 + 1 g impuritati
mp = 5g
m.ip = 5+1 = 6g
p% = mpx100/m.ip = 5x100/6 = 83,33%
dac in 6g proba .................. 5g bicarbonat
50g proba ............ m = 41,67g NaHCO3
41,67g m g
NaHCO3 + HCl --> NaCl + H2O + CO2
69 36,5
=> m = 41,67x36,5/69 = 22,05 g HCl