When a saturated solution of NH4Br dissolved in 100 grams of water is cooled from 60°C to 30°C, how much NH4Br will precipitate?
1 answer:
Answer:
Explanation:
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In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:
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Answer:
- The first picture attached is the diagram that accompanies the question.
- The<u> second picture attached</u> is the diagram with the answer.
Explanation:
In the box on the left there are 8 Cl⁻ ions and 8 Na⁺ ions.
The dissociaton equation for NaCl(aq) is:
- NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)
The dissociation equation for CaCl₂ (aq) is:
- CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)
A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.
Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half) and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).
Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.
Given parameters:
Mass of KClO₃ = 125g
Unknown:
Total mass of the products = ?
When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;
2KClO₃ → 2KCl + 3O₂
All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.
The total mass of the products must give us 125g according to this law of conservation of matter.
Now to find the masses of each product,
- Find the number of moles of the given reactant:
Number of moles =
molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol
So number of moles of KClO₃ = = 1.02moles
2. Now, using this number of moles, find the number of moles of the products using this value;
2 moles of KClO₃ produced 2 moles of KCl
1.02 moles of KClO₃ will also produce 1.02moles of KCl
2 moles of KClO₃ produced 3 moles of O₂
1.02 moles of KClO₃ will produce mole = 1.53 moles of O₂
3. Now find the masses of each product;
Mass = number of moles x molar mass
molar mass of KCl = 39 + 35.5 = 74.5g/mol
molar mass of O₂ = 16 x 2 = 32g/mol
Mass of KCl = 74.5 x 1.02 = 75.99g
Mass of O₂ = 32 x 1.53 = 48.96g
Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g
= 124.95g
This value is approximately the same as that of mass of KClO₃