A compound is composed of only C, H, and O. The combustion of a 0.519-g sample of the compound yields 1.24 g of CO2 and 0.255 g of H2O. What is the empirical formula of the compound

C4H4O

Explanation:

1.24 g x 1mol/(44)g = 0.03mol C;

Mass of C: 0.03mol x 12.g/1mol

= 0.36g of C

mol of H: 0.255g x 1mol /(18) x2= 0.028mol

Mass of H; = 0.028g × 1 of H

Mass of O: 0.519 g – (0.36g +0.028g)= 0.131g;

0.131g/16 =0.008 mol

Divide by the smallest number of moles;

0.028/0.008 mol = 3.5; 0.03/0.008 mol =3.75, 0.008 mol/0.008 mol =1

C4H4O

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the correct answer to this question is 2