How many moles of bromine gas are produced when fluorine reacts with 347.5 g of potassium
bromide?

1.46 mol

Explanation:

Given data:

Moles of bromine gas produced = ?

Mass of potassium bromide = 347.5 g

Solution:

Chemical equation:

2KBr  +  F₂   →  2KF + Br₂

Number of moles of KBr:

Number of moles = mass/ molar mass

Number of moles = 347.5 g/ 119.0 g/mol

Number of moles = 2.92 mol

Now we will compare the moles of KBr and bromine.

                  KBr       :     Br₂

                    2         :      1

                   2.92    :       1/2×2.92 = 1.46 mol

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