When CO2(g) is put in a sealed container at 730 K and a pressure of 10.0 atm and is heated to 1420 K , the pressure rises to 24.1 atm . Some of the CO2 decomposes to CO and O2. Calculate the mole percent of CO2 that decomposes.
Express your answer using two significant figures.

48%

Explanation:

Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:

P1T2 = P2T1

Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:

P1 = 10.0atm

T2 = 1420K

P2 = ?

T1 = 730K

P2 = 10.0atm*1420K / 730K

P2 = 19.45 atm

The CO2 reacts as follows:

2CO2 → 2CO+ O2

Where 2 moles of gas react producing 3 moles of gas

Assuming the 100% of CO2 react, the pressure will be:

19.45atm * (3mol / 2mol) = 29.175atm

As the pressure rises just to 24.1atm the moles that react are:

24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present

The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:

0.48mol / 1mol * 100 = 48%

h let me think

14,16 18 reacts with group