Question

Iron(III) oxide reacts with carbon monoxide according to the equation:

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

A reaction mixture initially contains 22.10 g Fe2O3 and 14.00 g CO.

Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains?

Express your answer to three significant figures.

Suppose that in an alternate universe, the possible values of l were the integer values from 0 to n(instead of 0 to n−1). Assuming no other differences from this universe, how many orbitals would exist in each of the following levels?

n = 4

Answer 1

Molar mass of Fe2O3 = molar mass of Fe*2 + molar mass of O*3

= 55.85 g/mole*2 + 16g/mole*3

= 111.69g/mole + 48g/mole

= 159.69g/mole

We have 22.1g of Fe2O3 so, moles of Fe2O3 = weight/molar mass

= 22.1g/ 159.69g/mole = 0.1384 moles

Also,

Molar mass of CO= molar mass of C + molar mass of O

= 12 g/mole + 16g/mole

= 28g/mole

We have 14g of CO so, moles of CO = weight/molar mass

= 14g/ 28g/mole = 0.5 moles

The balanced chemical reaction is -

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

So, each mole of Fe2O3 reacts with 3 moles of CO

But we have only  0.1384 moles of Fe2O3 so, they will react with  0.1384*3 moles of CO

0.1384*3 moles of CO = 0.4152 moles

So, after the reaction we will be left with excess of CO = 0.5 moles - 0.4152 moles

= 0.0848 moles

weight of 0.0848 moles of CO in g = molar mass*moles

= 28g/mole*0.0848 moles = 2.3744 g

So, excess of CO remaining after reaction = 2.3744 g