Answer:
-205.7kj
Explanation:
Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of the reaction.
The expression for enthalpy for the following reaction will be,
where,
n = number of moles
Now put all the given values in the above expression, we get:
Therefore, the enthalpy of the following reaction is, -205.7kj
Answer:
45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂
Explanation:
The balanced reaction is:
2 H₂ + O₂ → 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
Being the molar mass of the compounds:
then, by reaction stoichiometry, the following amounts of reactant and product mass participate:
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.
To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 4 g of H₂ react with 32 g of O₂, 36.8 g of H₂ with how much mass of O₂ will it react?
mass of O₂=294.4 grams
But 294.4 grams of O₂ are not available, 40.2 grams are available. Since you have less mass than you need to react with 36.8 grams of H₂, oxygen O₂ will be the limiting reagent.
Then you can apply the following rule of three: if by stoichiometry 32 grams of O₂ form 36 grams of H₂O, 40.2 grams of O₂ how much mass of H₂O will it form?
mass of H₂O= 45.225 grams
<u><em>45.225 grams of H₂O will be formed when 36.8 g H₂ is mixed with 40.2 g O₂</em></u>
Answer:
710,33 g NO2
Explanation:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 used to combust the octane
= 15.44 mol O2 used to form NO2
O2 + 2NO → 2NO2
(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710,33 g NO2