CHEMISTRY TEST 53 POINTS

How much heat should be transferred when 20.1 g of hydrogen bromide is formed during the reaction of hydrogen gas with liquid bromine?
72.80 kJ + H2 + Br2 → 2 HBr

9.1 KJ

Explanation:

First we must find the number of moles of HBr involved;

number of moles of HBr= mass of HBr/ molar mass of HBr

But molar mass of HBr= 80.91 g/mol

Given mass of HBr from the question= 20.1g

Hence;

Number of moles of HBr= 20.1 g/80.91g

Number of moles of HBr= 0.25 moles of HBr

If 2 moles of HBr has a heat of formation of 72.80KJ

Then 0.25 moles of HBr will have a heat of formation of 0.25× 72.80/2= 9.1 KJ

Therefore; 20.1 g of HBr will have a heat of formation of 9.1 KJ

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o