Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
the products formed from the reaction
all of the above
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted - 0.126 mol/L x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol
then 1000 mL contains - 0.00109 mol / 25 mL x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
The mass of oxygen reacted/required in this reaction is obtained as 48g.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.
Now, we have the reaction; S + O2 ------>SO2
If 1 mole of sulfur dioxide contains 22.4 L
x moles of sulfur dioxide contains 33.6L
x = 1.5 moles of sulfur dioxide.
Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.
Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g
Learn more anout stoichiometry: brainly.com/question/9743981