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.
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)
Saroj said:
4 years ago
Thanks for the good explanation @Rajesh.
(3)
Aditya sharma said:
7 years ago
I can't understand. Please explain the answer.
(3)
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)
Cindy said:
1 decade ago
Not clear! How?
(1)
Premnarayan said:
3 years ago
124 * 1.33/284 + 160*5.07/284 = 3.434.
(1)
Aruna said:
1 decade ago
Not clear please explain answer this question.
(1)
Prashant said:
1 decade ago
I can't understand this question. Please clear this question.
(1)
Vikram said:
1 decade 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
(1)
Joshua Benjamin bawa said:
8 years ago
What is a limiting agent please explain?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers