The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3
Mole fraction of Oxygen=0.381
Mole fraction of Oxygen= (range of moles of oxygen) ÷(general moles)
also, mole fraction of oxygen = (partial stress of oxygen) ÷ (total strain)
consequently , mole fraction of Oxygen= (2.31 atm)÷(2.31 atm + 3.75 atm)
= 0.381
The mole fraction may be calculated by means of dividing the variety of moles of 1 element of a solution by the entire quantity of moles of all the additives of a solution. It is cited that the sum of the mole fraction of all of the components inside the solution should be identical to 1.
Mole fraction is a unit of awareness. in the solution, the relative amount of solute and solvents are measured by way of the mole fraction and it's far represented through “X.” The mole fraction is the variety of moles of a selected aspect inside the answer divided by way of the entire range of moles in the given answer.
Mole fraction is the ratio between the moles of a constituent and the sum of moles of all ingredients in a mixture. Mass fraction is the ratio between the mass of a constituent and the full mass of a mixture.
The question is incomplete. Please read below to find the missing content.
Assuming that only the listed gases are present, what would the mole fraction of oxygen gas be for each of the following situations? A gas sample of 2.31 atm of oxygen gas and 3.75 atm of hydrogen gas react to form water vapor. Assume the volume of the container and the temperature inside the container does not change.
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All Elements are obtained in the periodic table & water is Not an element it is a compound.
Answer:
The answer to your question: 0.7 M
Explanation:
Data
V of KOH = 90 ml
[KOH] = ?
V H2SO4 = 21.2 ml
[H2SO4] = 1.5 M
2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
Molarity = moles / volume
moles of H₂SO₄ = (1.5) (21.2)
= 31.8
2 moles of KOH -------------- 1 mol of H₂SO₄
x -------------- 31.8 mol of H₂SO₄
x = (31.8)(2) / 1
x = 63.8 moles of KOH
Molarity = 63.8 / 90
= 0.7 M
