Answer:
0.67 L
Step-by-step explanation:
Given data:
Volume of carbon dioxide = ?
Mass of nitroglycerine = 2.5 g
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Solution:
Chemical equation:
4C₃H₅N₃O₉ → 12CO₂ + 10 H₂O + 6N₂ + O₂
Number of moles of nitroglycerine:
Number of moles = mass/ molar mass
Number of moles = 2.4 g/ 227.1 g/mol
Number of moles = 0.01 mol
Now we will compare the moles of nitroglycerine and carbon dioxide from balance chemical equation.
C₃H₅N₃O₉ : CO₂
4 : 12
0.01 : 12/4×0.01 = 0.03 mol
Volume of CO₂:
PV = nRT
V = nRT/P
V = 0.03 mol× 0.0821 atm.L/mol.K× 273.15 K / 1 atm
V = 0.67 L