One of the reactions for rusting iron is as follows: 4fe + 3o2 → 2fe2o3 (mm fe: 55.85 g/mol; mm o2=32 g/mol; mm fe2o3=159.70 g/mol) if 63.98 g of oxygen gas is completely consumed, how many moles of iron (iii) oxide are formed? a. 1.333 mol b. 3071 mol c. 2.999 mol d. 6812 mol

A. 1.333 mol

Explanation:

To calculate the moles, we use the equation:

As oxygen is completely consumed, it is the limiting reagent and will limit the formation of .

According to stochiometry,

3 mole of produces 2 moles of

Thus 2 moles of will produce= moles of

Thus the correct answer is 1.333 moles

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.