13. Combetion analysis of on unkan compuund contaising only carbon and hydrop pr
ID: 545029 • Letter: 1
Question
13. Combetion analysis of on unkan compuund contaising only carbon and hydrop prodaced 0.2845 g of CO2 sd 01451of HaO, Whet is the empirical formuls of the ompours c) Call o D)C3H2 14. Aocording to the following balanced reaction, bow mary mokes of NO une frened from &44 moles of Nop if here plenity ofwater pesenit A) 25.32 moles NO B) 12.66 moles NO C) 8.44 moles NO D) 5.63 moles NO E) 2.82 moles NO 15. Lithiom and nitrogen react to peoduce ithi nitride fHow many moles of lihium aitrids are produood whes 0450 mol of Trbim react in thia fashion? A) 0.150 B) 0.900 C10.0750 D) 1.35 E) 0225 16 Automotive air bags inflate when sodrem azide decomposes explosively to its con How many grams of sodiun acide aee required to prouce 3X0 g od nitrogen A) 1.77 B) 0.785 C) 766 D) 5t.1 E) 114.9Explanation / Answer
13)
Answer
B)C2H5
Explanation
Molar mass of CO2 = 44g/mol
Molar mass of C = 12g/mol
CO2 to C conversion factor = 0.2727
Mass of CO2 produced = 0.2845g
Mass of C = 0.2727 × 0.2845g = 0.07758g
No of mole of C = 0.07758g/12= 0.006465
Molar mass of H2O = 18g/mol
Molar mass of H = 1g/mol
H2O to H conversion factor = 2/18 = 0.1111
Mass of H2O produced = 0.1451g
Mass of H = 0.1111 × 0.1451 = 0.01612
if divide both no mole of C and No of mole of H by lowest mole ratio C:H is C1H2.5
if converting this for whole no
the ratio is
C2H5
Therefore,
Embirical formula is C2H5
14)
Answer
E) 2.82moles of NO
Explanation
stoichiometrically , 3moles of NO2 giving 1mole of NO
therefore, 8.44moles of NO2 give 8.44/3 = 2.82moles of NO
15)
Answer
A) 0.150
explanation
stoichiometrically , 2moles of Lithium Nitride is produced from 6moles of Li
therefore,
No of moles of Li3N produced by 0.450mole of Li = (2/6)×0.450 = 0.150
16)
Answer
D)51.1
Explanation
No of mole of N2 = 33g/28g/mol = 1.179mol
Stoichiometrically, 3moles of N2 produced by 2mole of NaN3
Therefore,
No of mole of NaN3 required ti produce 1.179moles of N2 = (2/3)×1.179= 0.786
Molar mass of NaN3 = 65g/mol
mass of NaN3 required = 65g/mol×0.786=51.1g
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