Vanuation Question
1 answer:
You might be interested in
Answer:
The percentage of the void space out of the total volume occupied is 93.11%
Explanation:
A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.
To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom
Let's calculate this one after the other.
For the hydrogen, formula for the volume will be
= 2 × 4/3 × π ×
where = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 ×
meters
Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31
we multiply this by the avogadro's number = 6.02 * 10^23
= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3
We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen
Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3
adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)
Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water
= (18g/mol)/(1 g/ml) = 18 ml/mol
Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml
Now, the percentage of void = volume of void/total volume * 100%
= 16.76/18 * 100% = 93.11%
Answer:
NO will be the limiting reagent.
Explanation:
The balanced equation is:
2 NO + 2 CO → N₂ + 2 CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- NO: 2 moles
- CO: 2 moles
- N₂: 1 mole
- CO₂: 2 moles
Being the molar mass of each compound:
- NO: 30 g/mole
- CO: 28 g/mole
- N₂: 28 g/mole
- CO₂: 44 g/mole
Then by stoichiometry the following quantities of mass participate in each reaction:
- NO: 2 moles* 30 g/mole= 60 g
- CO: 2 moles* 28 g/mole= 56 g
- N₂: 1 mole* 28 g/mole= 28 g
- CO₂: 2 moles* 44 g/mole= 88 g
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, you can use a simple rule of three as follows: If 56 grams of CO react with 60 grams of NO, 3 grams of CO react with how much mass of NO?
mass of NO= 3.21 grams
But 3.21 grams of NO are not available, 3 grams are available. Since you have less moles than you need to react with 3 grams of CO, <u><em>NO will be the limiting reagent. </em></u>
<u><em></em></u>