g A 0.4395 g sample of aluminum reacts according to our experiment to produce alum. 5.1629 g of dried alum crystals are recovered. What is the percent yield of the experiment?

92.75%

Explanation:

The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:

From above; we will see that 2 moles of Aluminium react with sulphuric acid and water to produce 2 moles o aluminium alum.

Therefore, the theoretical yield can be determined as:

= 4.789g of

To find the percent yield, we need to divide the actual yield by the theoretical yield and then multiply it with 100.

∴

percent yield = ( mass of alum(g)/theoretical yield(g) ) × 100

percent yield = ( 4.789g / 5.1629g ) × 100%

percent yield = 0.9275 × 100%

percent yield = 92.75%

Thus, the percent yield of the experiment 92.75%

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.

i dont know

explanation: