Blog Post 9

Blog Post 9 (Unit 9)

1. Did you finish and turn in your review to Mrs. Beau? (A simple yes or no is fine for this one). 

Yes. 

1. How is the molar mass calculated and why is it useful?

Molar mass is the complete overall mass of a compound elemental form. It is found through use of the periodic table elemental masses displayed, multiplied by the total number of each elemental component within the compound structure. Molar mass allows for the seperate elemental forms within compounds to be calculated and found without being known, 

1. Explain how molar mass can be used to determine the number of molecules in a sample of a compound if its mass and molecular formula are known.

By dividing the molar mass by the elements thought to be components within the formula, one can find the composition/how many of a specific element there are. 

An example of this, would be H2O. There are two Hydrogen atoms present, to every one oxygen bonded together. The molar mass shown for oxygen is 15.999, while Hydrogen's is 1.0008. As there are two of them, it would be multiplied by two, then added to the molar mass of the oxygen. 

1. Calculate the mass of SnF2 needed to form 10.0 g HF according to the formula: Sn + HF –> SnF2  + H2

Sn + 2HF --> SnF2 + H2  <-- Balanced 

10g HF (1mol HF / 2.006g HF) * (1mol SnF2 / 2mol HF) * (156.706g SnF2 / 1mol SnF2) = 39.2g SnF2

1. If you have 10.0 g of Sn also, how much SnF2 will be produced from reacting with the 10.0 g of HF? (Limiting reactant problem)

Sn + 2HF --> SnF2 + H2 <-- Balanced 

10.0g Sn (1mol Sn / 188.710g Sn) * (1mol SnF2 / 1mol HF) * (157.706g SnF2 / 1mol SnF) = 13.2g SnF2

10.0g HF (1mol HF / 20.006g HF) * (1mol SnF2 / 2mol HF) * (157.706g SnF2 / 1mol SnF) = 39.2g SnF2

The limiting reactant of this equation is Tin, 10 grams of Sn will create 13.2g of SnF2, and 10 grams of HF produces 39.2g of SnF2. 

1. Calculate the mass of Be needed to completely react with 18.9 g of N2 gas to produce Be3N2, which is the only product of the reaction.

N2 + 3Be --> Be3N2

18.9g N2 (1mol N2 / 28.14g N2) * (3mol Be / 1mol N2) * (9.012g Be / 1mol Be) = 18.2g Be

In order to completely react with 18.9 grams of N2 gas, 18.2 grams of Be are needed included in the reaction. 

1. Create a video response for describing the process of how you figured out the correct mass of sodium bicarbonate and volume of acetic acid to propel your baby bottle 6 meters. Go to Flipgrid to submit video. (This one is worth 20 points).