HELP
Phosphorus reacts with iodine to produce phosphorus(Ill) iodide:
P4 + 6I2 = 4PI3
What is the minimum mass of iodine required to produce 1 kg of phosphorus(Ill) iodide
when the phosphorus is in excess?
Data: molar mass of phosphorus(Il) iodide = 411.7 g mol
P4 + 6I2 = 4PI3
What is the minimum mass of iodine required to produce 1 kg of phosphorus(Ill) iodide
when the phosphorus is in excess?
Data: molar mass of phosphorus(Il) iodide = 411.7 g mol
924.7g
Original post by Allera
924.7g
could you show the working pls?
moles of PI3 = 1000/411.7
4PI3 = 2.43 moles
2.43/4 = 0.607
0.67 x 6 = 3.64 moles of 6I2
mass = 3.64 x 253.8
= 924g
4PI3 = 2.43 moles
2.43/4 = 0.607
0.67 x 6 = 3.64 moles of 6I2
mass = 3.64 x 253.8
= 924g
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