A chemist collects 2.94 L of carbon monoxide
(CO) gas at STP during an experiment. Explain
how she can determine the mass of gas that she
collected. Why is the mole important for this
calculation?

Respuesta :

Answer:

3.68g

Explanation:

The mole is central to all chemical calculations in quantitative chemistry. Without the mole, we cannot truly and accurately measure out a specified amount of substance or predict the quantity of the product likely to be obtained.

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

To determine the mass of gas;

  1. She must first find the number of moles of the gas.
  2. Using the equation;

                  Mass = number of moles x molar mass

  the mass can be determine;

 At stp;

      1 mole of any gas occupied a volume of 22.4L

      x mole of CO gas will occupy a volume of 2.94L

         22.4x = 2.94

               x = [tex]\frac{2.94}{22.4}[/tex] = 0.13mole

    Now to find the mass;

      Molar mass of CO = 12 + 16 = 28g/mole

     Mass = 0.13 x 28 = 3.68g

The mass of CO collected by chemist has been 3.67 grams. The moles have been important in calculation as it has been intermediate for the determination of mass of the gas.

The ideal gas has been considered of lacking the interatomic interactions and collisions.

The ideal 1 mole gas at standard temperature of 0 degree Celsius, and standard pressure of 1 atm has been considered of 22.4 L volume.

Computation of Mass at STP:

The CO has been considered as an ideal gas, The volume of gas at STP has been given as 2.94 L.

The moles of gas have been given as:

[tex]\rm 22.4\;L=1\;mol\;CO\\2.94\;L=\dfrac{1}{22.4}\;\times\;2.94\;mol\;CO\\2. 94\;L=0.13\;mol\;CO[/tex]

The moles of CO in the sample has been 0.13 mol.

The mass of gas has been given as:

[tex]\rm Mass=Moles\;\times\;Molar\;mass[/tex]

The mass of CO has been given as:

[tex]\rm Mass\;CO=0.13\;\times\;28.01\;g\\Mass\;CO=3.67\;g[/tex]

The mass of CO collected by chemist has been 3.67 grams. The moles have been important in calculation as it has been intermediate for the determination of mass of the gas.

For more information about ideal gas, refer to the link:

https://brainly.com/question/11676583