3 years ago

The decomposition of 11.0 g of fe2o3 results in

1 answer:

7 0

2Fe2O3=4Fe+3O2
1 mole Fe2O3=56*2+3*16=160g
2*160g Fe2O3.....4*56gFe.....6*16g O
11gFe2O3....x g Fe....y g O
x=11*4*56/(2*160)=7.7 g Fe
y=11*6*16/(2*160)=3.3g O

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