Answer:
Percent yield of PI3 = 95.4%
Explanation:
This is the reaction:
2P (s) + 3I2 (g) > 2PI3 (g)
Let's determine the moles of iodine that has reacted.
58.6 g / 253.8 g/mol = 0.231 mol
Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.
3 moles of I2 react to make 2 moles of PI3
0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3
As we have produced 0.147 moles let's determine the percent yield.
(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%
The equator is the parallel
I think 25m hopes dis helps
Answer:
The correct answer is - yes, 4.57 g of solute per 100 ml of solution
Explanation:
The correct answer is yes we can calculate the solubility of X in the water at 22.0°C. The salt will remain after the evaporate from the dissolved and cooled down at 26°C.
Then, the amount of solute dissolved in the 700 ml solution at 26°C is the weighed precipitate: 0.032 kg = 32 g.
Then solublity will be :
32. g solute / 700 ml solution = y / 100 ml solution
⇒ y = 32. g solute × 100 ml solution / 700 ml solution = 4.57 g.
Thus, the answer is 4.57 g of solute per 100 ml of solution.
