The bonds in the reactants of Figure 7-3 contained 372 kJ of chemical energy and the bonds in the products contained 350 kJ of c
hemical energy. What is the amount of energy change during the reaction (show your work for full credit)? Would this energy be absorbed or released? Explain how you know.
1 answer:
Answer: 22 kJ amount of energy is released in the following reaction.
Explanation: There are two types of reaction on the basis of amount of heat absorbed or released.
1. Endothermic reactions: These are the type of reactions in which reactants absorb heat to form the products. The energy of the reactants is less than the energy of the products.
2. Exothermic reactions: These are the type of reactions in which heat is released from the chemical reactions. The energy of the products is less than the reactants.
Sign convention for : This value is negative for exothermic reactions and positive for endothermic reactions.
For the given chemical reaction,
Energy of the products is less than the energy of the reactants, Hence, this reaction will be a type of exothermic reaction and energy will be released during this chemical change.
Amount of energy released = (350 - 372) kJ = -22kJ
Negative sign symbolizes the energy is being released. So, 22 kJ amount of energy is released in the following reaction.
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<h3>Answer:</h3>
The mass of excessive water (H₂O) is 40.815 kg
<h3>Explanation:</h3>The Equation for the reaction is;
Li₂O(s) + H₂O(l) → 2LiOH(s)
From the question;
Mass of water removed is 80.0 kg
Mass of available Li₂O is 65.0 kg
We are required to calculate the mass of excessive reagent.
<h3>Step 1: Calculating the number of moles of water to be removed</h3>Moles = Mass ÷ Molar mass
Molar mass of water = 18.02 g/mol
Mass of water = 80 kg (but 1000 g = 1kg)
= 80,000 g
Therefore;
= 4.44 × 10³ moles
<h3>Step 2: Moles of Li₂O available </h3>Moles = mass ÷ molar mass
Mass of Li₂O available = 65.0 kg or 65,000 g
Molar mass Li₂O = 29.88 g/mol
Moles of Li₂O = 65,000 g ÷ 29.88 g/mol
= 2.175 × 10³ moles Li₂O
<h3>Step 3: Mass of excess reagent </h3>From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)
1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH
The ratio of Li₂O to H₂O is 1:1
- Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
- However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
- This means, Li₂O is the limiting reactant while water is the excessive reagent.
Therefore:
Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles
= 2.265 × 10³ moles
Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol
= 4.0815 × 10⁴ g or
= 40.815 kg
Thus, the mass of excessive water is 40.815 kg