chemistry question

Question

For the following reaction, 19.4 grams of iron are allowed to react with 10.1 grams of oxygen gas .

iron(s) + oxygen(g)  iron(II) oxide(s)

What is the maximum mass of iron(II) oxide that can be formed?  grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?  grams

Answer 1

no of moles of Fe = 19.4/56 = 0.346

no of moles of oxygen = 10.1/32 = .315

2Fe + O₂ -→2 FeO

iron and oxygen react in 2:1 ratio

so 0.346 moles Fe will react with only 0.173 moles of oxygen.

Since Iron is present in less quantity in comparison to ratio it is the limiting reagent.

ratio between Fe : FeO is 1:1

therefore 0.346 moles of Fe will produce 0.346 moles of FeO

mass of FeO = moles X molar mass = 0.346 X (56+16) = 24.912 gms