Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 2 (Q.No. 32)
32.
1 kg of calcium carbide (CaC2) produces about 0.41 kg of acetylene gas on treatment with water. How many hours of service can be derived from 1 kg of calcium carbide in an acetylene lamp burning 35 litres of gas at NTP per hour ?
Discussion:
4 comments Page 1 of 1.
Jeetendra Kushwaha said:
6 years ago
= 22.4*0.41/(0.026*35) = 10.09 Hrs.
MD.Rabiul Alam said:
6 years ago
I do not understand it please describe details.
Mehtab Ali said:
3 years ago
The correct option is B 10 hrs.
CaC2+2H2O�\'Ca(OH)2+C2H2.
Number of moles of C2H2 = Given mass of C2H2/Molecular Mass of C2H2.
= 410g/26g = 15.7 mol.
According to the ideal gas equation,
PV = nRT
or
V=nRT/P= 15.7*0.0821*273.3/1 =350 L.
Number of hours of service T = Total amount of Acetylene produced /Amount of acetylene used per hour.
T= 350/35 = 10 hrs.
CaC2+2H2O�\'Ca(OH)2+C2H2.
Number of moles of C2H2 = Given mass of C2H2/Molecular Mass of C2H2.
= 410g/26g = 15.7 mol.
According to the ideal gas equation,
PV = nRT
or
V=nRT/P= 15.7*0.0821*273.3/1 =350 L.
Number of hours of service T = Total amount of Acetylene produced /Amount of acetylene used per hour.
T= 350/35 = 10 hrs.
(1)
Ramsajan said:
2 years ago
15*22.4=350.
Approximate 350/35 = 10hr.
Approximate 350/35 = 10hr.
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