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crimeas [40]
3 years ago
14

What type of reaction is represented by the following example? 2CO2 (g) + 4H2O (l) + 1452 kJ 2CH3OH (l) (g) + 3O2 (g) exothermic

endothermic
Chemistry
2 answers:
Elodia [21]3 years ago
8 0

endothermic  

I hope this helped :)

CaHeK987 [17]3 years ago
7 0

Answer : The given reaction is an example of endothermic reaction.

Explanation :

Endothermic reaction : It is a type of chemical reaction in which the energy is absorbed by the surroundings. In this reaction, the energy of reactants are less than the energy of products.  In this reaction, the energy is given to the system.

Exothermic reaction : It is a type of chemical reaction in which the energy is released into the surroundings. In this reaction, the energy of reactants are more than the energy of products. In this reaction, the energy is released from the system.

The given balanced chemical reaction is,

2CO_2(g)+4H_2O(l)+1452KJ\rightarrow 2CH_3OH(l)+3O_2(g)

In this reaction, the 1452 KJ energy is given to the system. So, the given reaction is an example of endothermic reaction.

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Explanation:

More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.

Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.

Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

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Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose
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Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

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Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624  = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

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