<span>Sucrose </span>is an example of a polymer separate but equal means that both races get the same thing they just dont get it at the same time.<span />
Answer:
[HOCH₂CH₂OH] = 24.1 m
Explanation:
Ethylene glycol → HOCH₂CH₂OH
60% by mass means that 60 g of ethylene glycol are contained in 100 g of solution.
Solution mass = Solute mass + Solvent mass
100 g = 60 g + Solvent mass
Solvent mass = 40 g
Molality are the moles of solute contained in 1kg of solvent.
We determine the moles of solute → 60 g . 1mol/62 g = 0.967 moles
We convert the mass of solvent from g to kg → 40 g . 1kg/1000 g = 0.04 kg
Molality → 0.967 mol / 0.04 kg = 24.1 m
Number of moles of sodium reacted is 2 moles. 0.04598 Kg is the mass of sodium that reacted.
Mole fraction is 1 mole ratio is 1:1
Explanation:
Balanced chemical reaction:
2 Na + 2H20 ⇒ 2NaOH + H2 (gas)
atomic mass of sodium = 22.99 g/mol
from the reaction it can be seen that 2 moles of sodium reacted to form 2 moles of sodium hydroxide soluition.
The mass of sodium reacted can be calculated from the formula:
mass = atomic mass of one mole x number of moles
mass = 22.99 x 2
= 45.98 grams is the mass of sodium, to convert it into kilograms it is divided by 1000
= 0.04598 Kg is the mass of sodium that reacted.
Mole fraction formula is:
= mole fraction ( solute Na = 2 moles, solution is NaOH =2 moles)
putting the values:
= 1 mole ratio is 1:1
Answer:
Explanation:
<u>1. Chemical quation</u>
The reaction of aluminium, sodium hydroxide and water is represented by the balanced chemical equation:
- 2Al(s) + 2NaOH(s) + 6H₂O(l) → 2Na[Al(OH)₄] (aq) + 3H₂(g) ↑
The coefficients of each reactant and product give the theoretical mole ratios.
To find the limiting reactant you compare the theoretical ratios with the ratio of the available substaces.
<u>2. Theoretical mole ratio:</u>
- 2 mol Al : 2 mol NaOH : 6 mol H₂O
Equivalent to
- 1 mol Al : 1 mol NaOH : 3 mol H₂O
<u>3. Actual ratio</u>
a) Convert each mass to number of moles
Formula:
- number of moles = mass in grams / molar mass
Al:
- molar mass = atomic mass = 26.982g/mol
- number of moles = 51.0g / 26.982g/mol = 1.89 mol
NaOH:
- number of moles = 84.1g / 39.997g/mol = 2.10 mol
H₂O:
- number of moles = 25.0g / 18.015g/mol = 1.39 mol
Divide all the mole amounts by the least number:
- Al: 1.89/1.39 = 1.36
- NaOH: 2.10 = 1.52
- H₂O: 1.39 = 1.00
- 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O
<u>4. Comparison</u>
<u />
Theoretical ratio:
- 1 mol Al : 1 mol NaOH : 3 mol H₂O
Actual ratio:
- 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O
Multiply by 3:
- 4.08 mol Al : 4.56 mol NaOH : 3.00 mol H₂O
Now, yo can see that the first two are in excess with respect the third one, making that the water consumes first, before any of the other two consumes. Therefore, the limiting reactant is the water.
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