Answer to Question #65182 in General Chemistry for Shayla

Question

if i have 5.6l of o2 at stp and 3.45 grams of fe, how many grams of iron 3 oxide would be produced

Answer 1

Solution.
4Fe + 3O2  2Fe2O3
n(O2) = 5.6/22.4 = 0.25 mole
n(Fe) = 3.45/56 = 0.06 mole
4 mole Fe ‒ 3 mole O2
x mole Fe ‒ 0.25 mole O2
x = 0.33 mole Fe > 0.06 mole
Fe ‒ limiting reagent
3.45 g Fe ‒ x g Fe2O3
(456) g Fe ‒ (2(562+163)) = 320 g Fe2O3
x = 3.45320/224 = 4.93 g Fe2O3

Answer: m(Fe2O3) = 4.93 g