Calculate the number of grams Fe2O3 produced from 8.93 g 02.
4 Fe +302 → 2 Fe2O3

Hey there!:

Molar mass :

Fe₂O₃ = 159.69 g/mol

O₂ = 31.99 g/mol

By reaction stochiometry  :

4 Fe + 3 O₂ = 2 Fe₂O₃

3 * (31.99 g O₂ )    2 * ( 159.69 ) g Fe₂O₃

8.93 g O₂ ( mass O₂)

mass O₂ =  ( 8.93 * 2 * 159.69 ) /  ( 3 * 31.99 )

mass O₂ = 2852.0634 / 95.97

mass O₂ = 29.71 g Of Fe₂O₃

Hope this helps!

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.