Examine the reaction equation.
1 answer:
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Answer:
70mol
Explanation:
The equation of the reaction is given as:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Given parameters:
Number of moles of acetylene = 35.0mol
Number of moles of oxygen in the tank = 84.0mol
Unknown:
Number of moles of CO₂ produced = 35.0mol
Solution:
From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.
To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:
From the equation:
2 moles of acetylene produced 4 moles of CO₂
∴ 35.0 mol of acetylene would produced:
= 70mol
Answer:
Therefore = -1835 KJ
Explanation:
Enthalpy is denoted by H.
Enthalpy: Total heat change in a chemical reaction is called enthalpy.
The change of entalpy of a reaction is denoted by
Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.
g= gas
S= solid
P₄(g)+10Cl₂(g)→ 4Cl₅(s) =?
PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) = +157KJ
P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) = -1207 KJ
If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.
We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.
PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) = -157KJ
Multiplying 4 with equation (3)
4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) =4×( -157)KJ= -628 KJ
Adding equation (2) and (4) we get
P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) =( -1207-628)KJ
⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) = - 1835KJ
⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) = -1835 KJ
Therefore = -1835 KJ