The formation of rust (Fe2O3) form solid iron and oxygen gas releases 1.7*10^3 k

Question

The formation of rust (Fe2O3) form solid iron and oxygen gas releases 1.7*10^3 kJ; 4Fe(s)+3O2(g)-->2Fe2O3(s) DeltaH + 1.7*10^3kJ. a) How many kJ are released when 2.00 g of Fe reacts? b) How many grams of rust form when 150 kcal are released?

Explanation / Answer

4Fe(s)+3O2(g)-->2Fe2O3(s) DeltaH + 1.7*10^3kJ dividing both side by 4 Fe(s)+3/4O2(g)-->1/2 Fe2O3(s) DeltaH =425 kj/mol a) How many kJ are released when 2.00 g of Fe reacts? so 1 mol Fe produces 425 kj 2.0 mol Fe*(425 Kj/1 mol Fe) 850 KJ b) How many grams of rust form when 150 kcal are released? 150 kcal = 627.6 Kj 627.6 kj * (

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