Answer:
ΔH°r = -483.64 kJ
Explanation:
Let's consider the following balanced equation.
2 H₂(g) + O₂(g) ⇒ 2 H₂O(g)
We can calculate the standard enthalpy change of the reaction (ΔH°r) using the following expression.
ΔH°r = ∑ΔH°f(p) × np - ∑ΔH°f(r) × nr
where
ΔH°f: standard heat of formation
n: moles
p: products
r: reactants
ΔH°r = ΔH°f(H₂O(g)) × 2 mol - ΔH°f(H₂(g)) × 2 mol - ΔH°f(O₂(g)) × 1 mol
ΔH°r = (-241.82 kJ/mol) × 2 mol - 0 kJ/mol × 2 mol - 0 kJ/mol × 1 mol
ΔH°r = -483.64 kJ
Answer:
84.3 g of nitrogen triiodide is the theoretical yield.
Explanation:
Hello there!
In this case, according to the chemical reaction:

It is possible to compute the theoretical yield of nitrogen triiodide by each reactant via stoichiometry as shown below:

Therefore, we infer that the smallest amount is the correct theoretical yield as it comes from the limiting reactant, in this case, diatomic iodine as it yields 84.3 g (three significant figures) of nitrogen triiodide as the theoretical yield; incidentally, nitrogen acts as the excess reactant.
Best regards!
I think it is b sorry if wrong
<h2>see in the attachment hope it helps you</h2>