General Knowledge - Chemistry - Discussion
Discussion Forum : Chemistry - Section 1 (Q.No. 16)
16.
The mass of P4O10 that will be obtained from the reaction of 1.33 gram of P4 and 5.07 of oxygen is
Discussion:
27 comments Page 2 of 3.
Joshua Benjamin bawa said:
8 years ago
What is a limiting agent please explain?
Ram said:
9 years ago
What is a limiting reagent?
Jatin said:
9 years ago
Quick formula. 1.33+5.07 = 6.4.
They combined producing another element. s6.4/2 = 3.2~3.05.
They combined producing another element. s6.4/2 = 3.2~3.05.
Udayanidhi said:
9 years ago
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.
(128g) (160g) (288g).
P4 is the limiting reagent.
128g of P4 -> 288g of P4O10.
1.33g of P4 -> (288/128) * 1.33g -> 3.05g.
Ayushi said:
9 years ago
Please explain the easy way to solve the solution.
Jahan said:
10 years ago
P4 + 5O2 = P4O10.
128 160 288.
128 g P4 produces 288g P4O10.
1.33 g P4 "(288/128)*1.33.
= 2.99 g~3.05 g.
128 160 288.
128 g P4 produces 288g P4O10.
1.33 g P4 "(288/128)*1.33.
= 2.99 g~3.05 g.
Ranjeet Yadav said:
1 decade ago
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.
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.
Awanish nishad (NIT Jalandhar) said:
1 decade ago
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.
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.
Faruque Abdullah said:
1 decade ago
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.
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.
Rajesh said:
1 decade ago
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.
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.
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