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1 answer:
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Answer:
- About 18 g of NH₄Cl will precipitate.
Explanation:
The <em>table G</em> is the graph of the solubility curves for several solutes which is attached.
The second picture identifies the solubilities for the NH₄Cl at 50ºC and 10ºC.
The solubility of NH₄Cl at 50ºC is about 52 g/ 100 g of water.
The solubility of NH₄Cl at 10ºC is about 34 g / 100 g of water.
Then, at 50ºC 100 g of water saturated with NH₄Cl contains about 52 g of NH₄Cl and 100 g of water saturated with NH₄Cl contains 34 g of NH₄Cl.
The difference, 52g - 34 g of NH₄Cl shall precipitate:
52 g - 34 g = 18g ← answer
Answer:
4.18 g
Explanation:
The formula for the calculation of moles is shown below:
Given: For Li
Given mass = 2.50 g
Molar mass of Li = 6.94 g/mol
<u>Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles</u>
Given: For
Given mass = 2.50 g
Molar mass of = 28.02 g/mol
<u>Moles of = 2.50 g / 28.02 g/mol = 0.08924 moles</u>
According to the given reaction:
6 moles of Li react with 1 mole of
1 mole of Li react with 1/6 mole of
0.3602 mole of Li react with mole of
Moles of that will react = 0.06 moles
Available moles of = 0.08924 moles
is in large excess. (0.08924 > 0.06)
Limiting reagent is the one which is present in small amount. Thus,
Li is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
6 moles of Li gives 2 mole of
1 mole of Li gives 2/6 mole of
0.3602 mole of Li react with mole of
Moles of = 0.12
Molar mass of = 34.83 g/mol
Mass of = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g
<u>Theoretical yield = 4.18 g</u>