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Virty [35]
2 years ago
10

Which of the following methods for increasing the rate of chemical reaction increases the effectiveness of the interactions betw

een reactants, instead of just causing the reactant molecules to bump into each other more frequently
Increasing the temperature of the reactants
Increasing the surface area of the reactants
Adding a catalyst to the reactants
Increasing the concentration of the reactants
Chemistry
1 answer:
denpristay [2]2 years ago
3 0

Answer:

Increasing the surface area of the reactants

Explanation:

An increase in surface area of the reactant will always cause an increase in the rate of reaction. This is so because, an increase in the surface area of the reactant will cause the reactant particles to collide effectively thereby bringing about an increase in the reaction rate.

Collision theory suggests that for a reaction to occur, the reactant particles must collide with the right orientation. As the surface area of the reactants are increased, the reactants particles collide more with the right orientation bringing about definite increase in the rate of reaction.

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16 grams of propane, C 3 H 8 and 20 grams of oxygen, O 2 are reacted to produce carbon dioxide and water. Calculate the volume o
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The volume of CO₂ produced is 8.4 L

Explanation:

The mass of propane in the reaction C₃H₈ = 20 grams

The mass of, oxygen, O₂ in the reaction = 20 grams

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The balanced equation of the (combustion) reaction can be presented as follows;

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (g)

Therefore, one mole of propane, C₃H₈, reacts with five moles of oxygen, O₂, to produce three moles of carbon dioxide, CO₂, and four moles of water molecules, H₂O, as steam

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The number of moles of propane in 16 grams of propane = 16/44.1 ≈ 0.3628 moles

The molar mass of oxygen, O₂ = 32.0 g/mol

The number of moles of oxygen in 20 grams of propane = 20/32 ≈ 0.625 moles

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Given that 1 mole of C₃H₈ reacts with 5 moles of O₂

1 mole of O₂ will react with 1/5 moles of C₃H₈

0.625 moles of O₂ will react with 0.625/5 = 0.125 moles of C₃H₈ to produce 3 × 0.125 = 0.375 moles of CO₂

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0.375 mole of CO₂ will occupy 0.375 × 22.4 L = 8.4 L

Therefore, the volume of CO₂ produced = The volume occupied by the 0.375 moles of CO₂ = 8.4 L.

5 0
2 years ago
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