Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
The complete balanced chemical reaction is written as:
AgNO3 + KCl ---> AgCl
+ KNO3
where AgCl is our
precipitate
So calculating for moles
of AgCl produced: MM AgCl = 143.5 g/mol
moles AgCl = 0.326 g /
(143.5 g/mol) = 2.27 x 10^-3 mol
we see that there is 1
mole of Ag per 1 mole of AgCl so:
moles Ag = 2.27 x 10^-3
mol
The molarity is simply
the ratio of number of moles over volume in Liters, therefore:
Molarity = 2.27 x 10^-3
mol / 0.0977 L
<span>Molarity = 0.0233 M</span>
Answer:
The masses of the reactants and products are equal.
Explanation:
Hope this helps.
