Answer:
Explanation:
You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.
The formula is
The equation for the reaction is
C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)
ΔS°/J·K⁻¹mol⁻¹ 219.5 205.0 213.6 69.9
The answer is likely to be A
Answer:
1,31÷2 =10,11
Explanation:
c10h22+31÷2o2=10co2+11h2o
The balanced reaction
is:
4NH3 + 3O2 --> 2N2 + 6H2O
<span>We
are given the amount of reactants to be used for the reaction. This
will be the starting point of our calculation.</span>
83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2
2.81 moles of NH3
From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.
<span>2.62 mol O2</span><span> (6 mol H2O / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.
