reactia chimica:
3Fe +4H2O->Fe3O4 + 4H2
Voi lucra in moli....nH2=m/M=178,6/2=89,3 moli
a)
3 moli Fe......4moliH2
xmoli Fe........89,3moli H2
x=89,3*3/4=66,975 moli Fe
mFe=n*M=66,975*56=3750,6grame Fe(3,75kg)
4moli H2O......4moli H2
ymoli H2O......89,3moli H2
y=89,3moli H2O
mH2O=n*M=89,3*18=1607,4grame H2O(1,6kg)
b)calculam masa molara de Fe3O4
M=3*56+4*16=232g/mol...asadar:
in 232g Fe3O4.....168g Fe.....64g O
in 100g Fe3O4......p1%........p2%
p1%=168*100/232=72,41%Fe
p2%=64*100/232=27,59%O
