Question

For the following reaction, 18.2 grams of iron are allowed to react with with 9.56 grams of oxygen gas.

I know this is multiple questions but it is in the same question on my HW and i'm so confused

What is the maximum amount of iron(II) oxide that can be formed? And What is the FORMULA for the limiting reagent? And What amount of the excess reagent remains after the reaction is complete?

Answer 1

Solution :-

18.2 g Fe and 9.56 g O2

mass of product = ?

formula of limiting reactant =?

mass of excess reactant = ?

Balanced reaction equation is as follows

4Fe(s) +3O2(g) ------ > 2 Fe2O3(s)

Using the mole ratio we can calculate the amount of product formed from the given amount of the reactants

moles = mass / molar mass

moles of Fe = 18.2 g / 55.845 g per mol = 0.3259 mol Fe

moles of O2 = 9.56 g / 32 g per mol = 0.29875 mol O2

Now lets calculate the moles of O2 needed to react with moles of Fe using the mole ratio

0.3259 mol Fe * 3 mol O2 / 4 mol Fe =0.2444 mol O2

moles of O2 needed for the reaction are less than moles of O2 present

therefore the limiitng reactant is the Fe and excess reactant is O2

now lets find the amoutn of O2 remain after the reaction

moles of O2 remain = 0.29875 mol - 0.2444 mol = 0.05435 mol O2

lets find mass of excess O2 remain

mass = moles * molar mass

mass of O2 remain = 0.05435 mol O2 * 32 g per mol = 1.739 g O2

So amoutn of O2 remain in excess is 1.739 g

Now lwts calculate the mass of the product Iron (lll) oxide (Fe2O3) produced

Fe is the limiting reactant therefore using the mole ratio of the Fe and Fe2O3 lets calculate the mass of the Fe2O3 that can be formed

(0.3259 mol Fe * 2 mol Fe2O3 / 4 mol Fe)*(159.69 g / 1 mol Fe2O3)= 26.0 g Fe2O3

Therefore the amount of the product Iron (lll) oxide (Fe2O3) that can be formed from the reaction is 26.0 g Fe2O3