What volume of o2 is needed to react fully with 720. Ml of nh3
1 answer:
Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,
From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with
L of oxygen gas.
So, 0.720 L of ammonia will react with:
Therefore, the volume of oxygen required is 900 mL.
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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