Phosphorus reacts with iodine to produce phosphorus(III) iodide:
P4 + 6I2 → 4PI3
What is the minimum mass of iodine required to produce 1 kg of phosphorus(III) iodide
when the phosphorus is in excess?
Data: molar mass of phosphorus(III) iodide = 411.7gmol−1
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Edexcel Chemistry Question 2018 Jan watch
- Thread Starter
- 02-03-2018 11:34
- 02-03-2018 12:46
Calculate the moles of PI3. The moles of iodine is 6/4 that amount.
Then calculate the molar mass of I2 (twice the mass of an iodine atom) and its mass.
The answer is 924.70 g = 0.925 kg (B)
Where did you find this anyway? Isn't the paper locked?
- 20-05-2018 01:01
4P + 6I2 -----> 4PI3
Okay so here, they're asking for the mass of iodine for 1000 g (1 kg) pf Phosphorus iodide.
You have the mass, the molar mass, so you get out the number of moles using number of mol = mass/molar mass. mass should be in grams of course.
1000/411.7 = 2.43 mol
but in the equation they have 4.
PI3: 4 ----> 2.43
I2: 6 ------> ?
? = (6 x 2.43)/4 = 3.64 mol
Now you have Iodine's number of mol, you need the molar mass to get its mass, its molar mass is 254(since it's I2 molecule, not one atom)
3.64 x 254 = 924.56 grams
if you want it in kg, divide by 1000 you'll get 0.9245.