If 4.9 kg of CO2 are produced during a combustion reaction, how many molecules of CO2 would be produced?

Considering the definition of avogadro's number, 6.707×10²⁵ molecules of CO₂ would be produced.

Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole.

Avogadro's number applies to any substance.

First, you must know the number of moles that contain 4.9 kg= 4900 g (being 1 kg=1000 g) of CO₂. For that, you must know that the molar mass of CO2, that is, the mass that contains one mole of the compound, is 44 g/mole.

Then, the number of moles of CO₂ can be calculated as:

So, you can apply the following rule of three: if by definition of Avogadro's Number 6.023×10²³ molecules are contained in 1 mole of CO₂, then how many molecules of CO₂ are composed of 111.36 moles?

amount of molecules of CO₂= (6.023×10²³ molecules × 111.36 moles)÷ 1 mole

amount of molecules of CO₂= 6.707×10²⁵ molecules

Finally, 6.707×10²⁵ molecules of CO₂ would be produced.

Learn more about Avogadro's Number:

6.7 x 10²⁶molecules

Explanation:

Given parameters

Mass of CO₂  = 4.9kg  = 4900g

Unknown:

Number of molecules  = ?

Solution:

To find the number of molecules, we need to find the number of moles first.

 Number of moles  =

          Molar mass of CO₂  = 12 + 2(16)  = 44g/mol

   Number of moles  =  = 111.36mole

A mole of substance is the quantity of substance that contains the avogadro's number of particles.

       1 mole  = 6.02 x 10²³molecules

     111.36 moles  =   111.36 x 6.02 x 10²³molecules   = 6.7 x 10²⁶molecules

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%