Answer:
1. 4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)
2. 238 g of C₃H₅(NO₃)₃ has been reacted.
Step-by-step explanation:
This is the chemical reaction:
4C₃H₅(NO₃)₃ (l) → 12CO₂(g) + 10H₂O(g) + O₂(g) + 6N₂(g)
For the second part, let's apply the Ideal Gases Law to find out the moles of CO₂ that were produced.
P . V = n . R . T
1 atm . 69L = n . 0.082 L.atm/mol.K . 268K
(1 atm . 69L) / (0.082 L.atm/mol.K . 268K) = n → 3.14 moles
In the equation, ratio between nitroglycerin and CO₂ is 12:4.
12 moles of CO₂ were produced by 4 moles of C₃H₅(NO₃)₃
Then, 3.14 moles of CO₂ would have been produced by (3.14 .4) / 12 = 1.04 moles of C₃H₅(NO₃)₃
Let's convert the moles to mass, to find out the mass of nitroglycerin that must have reacted (mol . molar mass)
1.04 mol . 227.08 g/mol = 238 g