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Fe2O3 + H2 --> Fe + H2O

Awhat mass of hydrogen gas must be consumed to produe 10.0 g of iron metal?

Bwhat mass of iron(lll) oxide, Fe2O3, must be consumed to prepare 2.50g of iron metal

Verified Answer :

Hippalectryon
A10.0g of Iron is \(\frac{10.0}{55.8}=179mmol\) of Iron.

One mole of dihydrogen is required to form one mole of iron, hence you'll need \(0.179*1.00=0.179g\) of dihydrogen.

B) Likewise : 2.50g of Iron is \(\frac{2.50}{55.8}=44.8mmol\) of Iron, hence we'll need \(44.8*(3*16+2*55.8)=7.15g\) of Fe2O3.

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