Select the option that correctly completes this statement: If the reaction takes place at a constant volume in a thermally insul
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Answer : The correct option is, (A)
Explanation :
Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.
We can determine the van't Hoff factor by the association and dissociation of the compound.
(A)
It is an electrolyte that dissociates to give aluminum ion and chloride ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 3 = 4
(B)
It is an electrolyte that dissociates to give potassium ion and iodide ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 1 = 2
(C)
It is an electrolyte that dissociates to give calcium ion and chloride ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 2 = 3
(D)
It is an electrolyte that dissociates to give magnesium ion and sulfate ion.
The dissociation of will be,
So, Van't Hoff factor = Number of solute particles = = 1 + 1 = 2
(E) Non-electrolyte
The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.
Hence, the highest van't Hoff factor of solute is,
Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CH₄ + 2 O₂ → CO₂ + 2 H₂O
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:
- CH₄: 1 mole
- O₂: 2 moles
- CO₂: 1 mole
- H₂O: 2 moles
The molar mass of the compounds is:
- CH₄: 16 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CH₄: 1 mole ×16 g/mole= 16 grams
- O₂: 2 moles ×32 g/mole= 64 grams
- CO₂: 1 mole ×44 g/mole= 44 grams
- H₂O: 2 moles ×18 g/mole=36 grams
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.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?
mass of O₂= 34 grams
But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.
<h3>Mass of CO₂ formed</h3>The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?
<u><em>mass of CO₂= 10.93 grams</em></u>
Then, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.
Learn more about the reaction stoichiometry:
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