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
Answer:
Explanation:
Hello,
In this case, the only source of hydrogen is in the 6 molecules of water, therefore, the atoms of hydrogen, by applying stoichiometry with the Avogadro's number is:
Best regards.
Answer:
16.9g
Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2
Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.
Then to calculate the theoretical yield you need to compare moles of the reactants:
m(Cu)=5g
M(Cu)=63.55
n(Cu)=5/63.55=0.0787
By comparing coefficients you require twice as much silver: 0.157mol
n(Ag)=0.157
M(Ag)=107.86
m(Ag)=0.157x107.86=16.9g
Hence, the theoretical yield of this reaction would be 16.9g
Answer:
Explanation:
A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products . A product is a substance that is present at the end of a chemical reaction.
h2 + 02 = h2o
<h2 />
