Problem 3.70 Detonation of nitroglycerin proceeds as follows: 4CaHs N309 ()-+12C

Question

Problem 3.70 Detonation of nitroglycerin proceeds as follows: 4CaHs N309 ()-+12CO2 (g) 6N2 (g) O2 (g) 10H20lg) previous l 19 of 20 l nex Part A 1,592g/mL) is a sample containing 300 ml, of nitroglycerin (density detonated, how many total moles of gas are produced? ri- 0.0761 mol Submit My Answers Give Up Incorrect: Try Again Part B each mole of gas occupies 59 Lunder the conditions of the explosion, how many liers of gas are produced? Express your answer using two significant figures. v. 4.489 Submit My Answers Give Up

Explanation / Answer

density= mass/volume
moles=mass/ molar mass

nitroglycerin mass= density *volume= 1.592*3 = 4.776 g

moles of nitroglycerin = 4.776 /227.0865 = 0.021

As per equation, 4 moles of nitroglycerin given 12 moles of CO2, 6 moles of N2, 1 mole of O2 and 10 moles of H2O all are gases

so, 0.021 moles of nitroglycerin given 0.063 moles of CO2, 0.0315 moles of N2, 0.00525 mole of O2 and 0.0525 moles of H2O
-------------------------------------------------------------------

Hence, for part A -> n= 0.063+0.0315+0.00525+0.0525 = 0.15225

---------------------------------------------------------------------
For part B , 1 mole occupies 59 L
hence, 0.15225 moles occupies 8.98275L

----------------------------------------------------------------------
FOr part C,
moles of N2 produced = 0.0315

mass of N2 = moles *molar mass = 0.0315*28.013 = 0.8824095

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.