Answer:
5.52 g
Explanation:
First we <u>convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 6.30 g NH₃ ÷ 17 g/mol = 0.370 mol NH₃
- 1.80 g O₂ ÷ 32 g/mol = 0.056 mol O₂
Now we <u>calculate with how many NH₃ moles would 0.056 O₂ moles react</u>, using the<em> stoichiometric coefficients</em>:
- 0.056 mol O₂ *
= 0.045 mol NH₃
As there more NH₃ moles than required, NH₃ is the excess reactant.
Then we calculate how many NH₃ moles remained without reacting:
- 0.370 mol NH₃ - 0.045 mol NH₃ = 0.325 mol NH₃
Finally we convert NH₃ moles into grams:
- 0.325 mol NH₃ * 17 g/mol = 5.52 g
Answer:
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Answer:
Kc = 0.075
Explanation:
The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:
α = x/M
x = M*α
x = 0.354M
For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:
2HI(g) ⇄ H₂(g) + I₂(g)
M 0 0 <em> Initial</em>
-0.354M +0.177M +0.177M <em>Reacts</em>
0.646M 0.177M 0.177M <em>Equilibrium</em>
The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):
![Kc = \frac{[H2]*[I2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BH2%5D%2A%5BI2%5D%7D%7B%5BHI%5D%5E2%7D)
Kc = 0.075
