-calculez NR.MOLI de metan, in c.n.
pV=niuRT --> niu= 2x55,39/0,082x(27+273)----> niu=4,5molCH4
-ECUATIA REACTIEI
1mol 1mol
CH4+O2--> CH2O +H2O
4,5mol............niu-=4,5mol metanal daca randamentul e 100% este niu teoretic; va rezulta insa mai putin
70/100= niu/4,5----> niu,practic=3,152 mol metanal
-calculez masa obtinuta: m= 3,152molX(12gC+2gH+16gO)/mol=94,5g
-aceasta trebuie sa fie masa dizolvata in solutia 40%
c/100= md/ms---< 40/100= 94,5/ms---> ms2 calculeaza !!!!!!