Consider a situation in which 161 g of p4 are exposed to 176 g of o2. a.)what is the maximum amount of p2o5 that can theoretically be made from 161 g of p4 and excess oxygen? (i got 2.60 mol) b.)what is the maximum amount of p2o5 that can theoretically be made from 176 g of o2 and excess phosphorus? (i got 2.20 mol) c.)now, determine how many moles of p2o5 are produced from the given amounts of phosphorus and oxygen. (i got 2.20mol/ theoretical yield) is what i need d.)what is the percent yield if the actual yield from this reaction is 206g?

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