If 11.5 grams of chlorine reacts with aluminum, how many grams of aluminum chloride will be formed according to the following reaction?

2 al + 3 cl2 à 2 alcl3

the balanced equation for the reaction is as follows

2 Al + 3 Cl₂ ---> 2 AlCl₃

molar ratio of Cl₂ to AlCl₃ is 3:2

mass of Cl₂ reacted is - 11.5 g

number of moles of Cl₂ reacted - 11.5 g / 71 g/mol = 0.162 mol

when 3 mol of Cl₂ reacts - 2 mol of AlCl₃ is formed

therefore when 0.162 mol of Cl₂ reacts - 2/3 x 0.162 mol = 0.108 mol of AlCl₃ is formed

the molar mass of AlCl₃ is 133 g/mol

mass of AlCl₃ formed is - 0.108 mol x 133 g/mol = 14.4 g

mass of AlCl₃ formed is 14.4 g

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.