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xgH2---80g Fe2O3
6g----160g
x=3g
potrzeba 3g wodoru czyli 5g wystarczy
mH₂ = 2u
mFe₂O₃ = 2*56u + 3*16u = 112 + 48 = 160u
3*2u H₂ --- 160u Fe₂O₃
x ---- 80g Fe₂O₃
6*80 = 160x
480 = 160x
x = 3 g H₂
Odp: 5 gram wodoru wystarczy