Question

What is the maximum mass of p2i4 that can be prepared from 8.07 g of p4o6 and 9.97 g of iodine according to the reaction 5p4o6 + 8i2 yields 4p2i4 + 3p4o10

Answer 1

@Rushwr

Answer 2

@Data_LG2

Answer 3

I think you have to convert the given masses of the reactants into moles first. Then using mole ratios, you can be able to determine which will yield the least amount of product, which will be your limiting reagent. The limiting reagent will determine how much of the product will be produced.

Answer 4

@owlet could you show me through a equation on what to do?

Answer 5

@ganeshie8

Answer 6

Step 1:

given mass of p4o6 x molar mass of p4o6 x (4 mole p2i4/ 5 p4o6 ) = ____ moles of p2i4
given mass of i2 x molar mass of i2 x ( 4 mole p2i4/ 8 mole i2 ) = _____ moles of p2i4

Step 2:
use the smallest amt of moles of p2i4 from step 1 to solve for the mass of p2i4

_____ moles of p2i4 x (molar mass of p2i4) = _____g of p2i4 --> this is the value that the question is asking.

Answer 7

Okay so 8.07 g * 219.8 * (4/5) ?

Answer 8

Is that right?

Answer 9

@Photon336

Answer 10

let's balance the reaction first

Answer 11

It's already balanced isn't it?

Answer 12

5P4O6 + 8I2 --> 4P2I4 + 3P4 O10

Answer 13

\[5P_{4}O_{6} + 8I_{2} --> 4P_{2}I_{4} + 3P_{4}O_{10}\]

Answer 14

Good

Answer 15

Hey can you find the molar masses of all these compounds first before we even do anything.

Answer 16

P4O6 = 219.8
I2 =126.9

Answer 17

@staldk3 the first thing we must do is convert both to moles.