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 1 of 3.
Premnarayan said:
3 years ago
124 * 1.33/284 + 160*5.07/284 = 3.434.
(1)
Saroj said:
4 years ago
Thanks for the good explanation @Rajesh.
(3)
Jagath said:
7 years ago
Thanks for explaining.
Aditya sharma said:
7 years ago
I can't understand. Please explain the answer.
(3)
Shiva said:
7 years ago
Why We using p-value (1.33) only in this calculation?
Why not using o value (5.07)?
Please explain me.
Why not using o value (5.07)?
Please explain me.
Nibedita said:
7 years ago
The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete.
Ajanaku opemipo said:
8 years ago
It is 3.05.
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.
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)
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)
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