General Knowledge - Chemistry - Discussion

Four mole phosphorus (4* mwt 31=124)equal to one mole of P4 (mwt 124) react with 5 moles of Oxygen molecule (5* mwt 32=160)to give one mole of Phosphorus Decaoxide (mwt 284)

124 gm P4 will react with 160 gm oxygen molcule (O2) to give 284 gm P4O10.

Therfore 1.33 gm P4 will react with 1.72 gm of O2 (1.33*32*5/124 = 1.72, here Oxgen is taken in excess i.e. 5.07 gm)) will give (1.33*284/124 = 3.05 gm of P4O10.
So the right answer is option [B]

@ Mehulkumar: Here you have to take molecular wt of oxygen molecule and not oxygen atom.

p4+5 O2 =P4O10.
124gm 160gm 284gm.

Moles of p4 consumed =1.33/124=0.0107258065.
Moles of o2 consumed=5.07/160=0.0316875.

Now for limiting reagent.
p4 is limiting.

Now,
1 mole of p4 produced 1 mole of p4O10
Then 0.0107258065 mole.......=0.0107258065 mole.

Now weight of p4o10 in gm=moles *molecular wt.
=0.0107258065*284=3.046129046.

Approximate answer 3.05.

P4 + 5 O2 P4O10.

Here atomic weight of phosphorus = 31 & for 4 molecules of P it becomes 124.

Also atomic weight of O is 16 & for 10 molecules it becomes 160.

Total atomic weight of product is 284.

As 124 gms P = 284 gms. P4O10.

Now for 1.33 gms= ?

For 1.33 gm = 284*1.33/124 = 3.0461 i.e. 3.05.

The limiting reagent (or limiting reactant) in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.

Balance chemical equation
P4 + 5 O2 -> P4O10
1 mol 5 mol 1 mol
124g 160g 284g

Here limiting reagent is P4.

124g of P4 produce P4O10 = 284g
So, 1.33g of P4 produce P4O10 = (284/124)*1.33 = 3.05g

Balance chemical equation.

P4 + 5 O2 -> P4O10,
1 mol 5 mol 1 mol.
124g 160g 284g.

Here limiting reagent is P4.
124g of P4 produce P4O10 = 284g.
So, 1.33g of P4 produce P4O10 = (284/124)*1.33 = 3.05g.

P4+5O2 = P4O10.
124 gm+ 160 gm = 284 gm.
284 gm produce from 124 gm.

1.33 gm will react with (284/124)*1.33 gm=3.046129032. That means 3.05. Therefore the right answer will B.

P4 + 5O2 -> P4O10 is the balanced equation.
(128g) (160g) (288g).

P4 is the limiting reagent.

128g of P4 -> 288g of P4O10.
1.33g of P4 -> (288/128) * 1.33g -> 3.05g.

P4=124gm,5O2=160gm.
p4+5o2=p4o10
160gm.
124gm+160gm=284gm.
284\124*1.33gm=2.2903.
2.2903 * 1.33=3.0461.
And 3.05gm is the answer.

P4+5O2 ->P4O10
124+5(16)=204
124gm Of P4 produce 204 gram P4O10
So,
1.33gm P4 produce (204/124)*1.33=2.18gm
So close answer is 2.04gm Option (A)