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.
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.
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.
Ayushi said:
9 years ago
Please explain the easy way to solve the solution.
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.
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.
Ram said:
9 years ago
What is a limiting reagent?
Joshua Benjamin bawa said:
8 years ago
What is a limiting agent please explain?
Amlesh kumar said:
8 years ago
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
160gm.
124gm+160gm=284gm.
284\124*1.33gm=2.2903.
2.2903 * 1.33=3.0461.
And 3.05gm is the answer.
(2)
Asutosh behera said:
8 years ago
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 + 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.
(5)
Noor said:
8 years ago
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.
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