How many grams are needed to prepare 0.289 liters of a 0.00300 M CU(NO3) solution

How many grams are needed to prepare 0.289 liters of a 0.00300 M CU(NO3) solution

2.54 mol CO2

Explanation:

To calculate the mol of carbon dioxide that will be generated by megajoule (MJ). First, we have to find the value of the formation enthalpy for the CO2 at standard conditions.

Remember that standard conditions are at 25°C and 1 atm.

Any physic chemistry book has the table of ΔHf°, I always use one name Levine's physic chemistry. You can trust its data.

Enthalpy of formation for CO2: -393.509 kJ/mol

As you can see this value means that for each 393.509 kJ liberate (that's indicated for the menus sing),  1 mol of CO2 is formed.

Let's write this using mathematic language:

393.509 kJ/ 1 mol CO2

The problem asks you to find how many mols are produced by 1 megajoule liberate, let's change megajoule to kilojoule.

1 megajoule (MJ) = 1000 kJ

1000 kJ x (1mol/ 393.509 kJ) = 2.54 mol CO2  

Remember don't use the negative value to make the calculus. The negative value just indicates that the heat is been released.

If you want to watch the mathematics details, go to the image attached.