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
Grams of Ca(NO₃)₂ produced : 0.985 g
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction
CaCl₂ + 2AgNO₃ → 2AgCl + Ca(NO₃)₂
MW AgNO₃ : 107.9+14+3.16=169.9
mol AgNO₃ :
mol ratio Ca(NO₃)₂ : AgNO₃ = 1 : 2, so mol Ca(NO₃)₂ :
MW Ca(NO₃)₂ : 40.1+2.14+6.16=164.1 g/mol
mass Ca(NO₃)₂ :
First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,
<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>
<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>
<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
Answer:
the systematic name is called Lithium sulfide.
