Question

In: Chemistry

A 52 g sample of iron reacts with 22 g of oxygen to form how many...

A 52 g sample of iron reacts with 22 g of oxygen to form how many grams of iron oxide?

Express your answer using two significant figures.

Solutions

Expert Solution

Answer – We are given, mass of Fe = 52 g , mass of O2 = 22 g

4 Fe + 3O2 ------> 2 Fe2O3

First we need to calculate the moles of each reactant

Moles of Fe = 52 g / 55.845 g.mol-1 = 0.931 moles

Moles of O2 = 22 g / 31.998 g.mol-1 = 0.687 moles

Now we need to calculate the limiting reactant

Moles of Fe2O3 from Fe

From the balanced equation

4 moles of Fe = 2 moles of Fe2O3

So, 0.931 moles of Fe = ? moles of Fe2O3

= 0.931 moles of Fe * 2 moles of Fe2O3 / 4 moles of Fe

= 0.466 moles of Fe2O3

Now moles of Fe2O3 from O2

From the balanced equation

3 moles of O2 = 2 moles of Fe2O3

So, 0.687 moles of O2 = ? moles of Fe2O3

= 0.687 moles of O2 * 2 moles of Fe2O3 / 3 moles of O2

= 0.458 moles of Fe2O3

So limiting reactant is O2, since the moles of Fe2O3 got lowest from the moles of O2.

So moles of Fe2O3 = 0.458 moles

Mass of Fe2O3 = 0.458 moles * 159.687 g/mol

                         = 73 g

So, a 52 g sample of iron reacts with 22 g of oxygen to form 73 grams of iron oxide


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