The rusting of iron is represented by the equation 4Fe + 3O2 --> 2Fe2O3. If you have a 3.5-mol sample of iron, how many grams of O2 are needed to rust the iron completely? 34.9 g
84.0 g
16.5 g
48.6 g

84.0 g

Balanced equation: 4Fe + 3O2 --> 2Fe2O3

3.5-mol sample of iron

Find moles of oxygen:

4Fe : 3O24 : 3(3.5/4)*3 = 2.625 moles

Find mass of O2

mass = moles * Mrmass = 2.625 * 32mass = 84 g

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts