Answer :
Total moles of gas = 0.1225
Volume of gas produced : 6.7375 L
mass of Nitrogen : 0.588 g
Further explanation
Given
2 ml of Nitroglycerin(ρ=1.592 g/ml)
Required
Total moles of gas
Solution
Nitroglycerin detonated ⇒ decomposition reaction
4C₃H₅N₃O₉(s)⇒ 6N₂(g)+12CO(g)+10H₂O(g)+7O₂(g)
mass of Nitroglycerin :
[tex]\tt mass=2~ml\times 1.592~g/ml=3.184~g[/tex]
moles of Nitroglycerin :
[tex]\tt moles=\dfrac{3.184}{227,0865~g/mol}=0.014[/tex]
Total moles of gas:
[tex]\tt \dfrac{6+12+10+7}{4}\times 0.014=0.1225[/tex]
Volume of gas produced :
[tex]\tt 0.1225\times 55=6.7375~L[/tex]
moles of Nitrogen :
[tex]\tt \dfrac{6}{4}\times 0.014=0.021[/tex]
mass of Nitrogen :
[tex]\tt 0.021\times 28=0.588~g[/tex]