Iron (III) nitrate reacts with enough sodium iodide to produce iron (II) iodide and other products. For the balanced equation shown below if the reaction of 241.86 grams of Fe(NO3)3 proceeds with a 90% yield, how many grams of iron (II) iodide, FeI2 would be produced?

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol