How many moles are in 25.0 grams of KMnO4?
1 answer:
Answer:- 0.158 moles
Solution:- Moles are calculated on dividing the grams by the molar mass.
25.0 grams of are given and we are asked to calculate the moles.
For this we need the molar mass so let's calculate it first.
Molar mass is the sum of atomic masses of all the atoms present in the molecule.
Molar mass =
= 158.036 gram per mol
Now, we will divide the given grams by molar mass to get the moles and the set is shown as:
=
So, there are 0.158 moles of .
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Answer:
ΔS = +541.3Jmol⁻¹K⁻¹
Explanation:
Given parameters:
Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹
Standard Entropy of C = 5.7Jmol⁻¹K⁻¹
Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹
Standard Entropy of CO = 198Jmol⁻¹K⁻¹
To find the entropy change of the reaction, we first write a balanced reaction equation:
Fe₂O₃ + 3C → 2Fe + 3CO
To calculate the entropy change of the reaction we simply use the equation below:
ΔS = ∑S - ∑S
Therefore:
ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1
ΔS = +541.3Jmol⁻¹K⁻¹