Answer:
1. 2 C₃H₅N₃O₉(l) ⇒ 3 N₂(g) + 1/2 O₂(g) + 5 H₂O(g) + 6 CO₂(g)
2. 146 g of nitroglycerin.
Explanation:
1. Write a balanced chemical equation, including physical state symbols, for the decomposition of liquid nitroglycerin ( C₃H₅N₃O₉) into gaseous dinitrogen, gaseous dioxygen, gaseous water and gaseous carbon dioxide.
The equation is:
C₃H₅N₃O₉(l) ⇒ N₂(g) + O₂(g) + H₂O(g) + CO₂(g)
Since atomicities in nitroglycerin are odd, it is easier to balance this equation by multiplying this compound by 2. The balanced equation is:
2 C₃H₅N₃O₉(l) ⇒ 3 N₂(g) + 1/2 O₂(g) + 5 H₂O(g) + 6 CO₂(g)
2. Suppose 41.0L of carbon dioxide gas are produced by this reaction, at a temperature of −14.0°C and pressure of exactly 1 atm. Calculate the mass of nitroglycerin that must have reacted.
First, we have to find the moles of COâ‚‚ using the ideal gas equation.
P.V = n . R . T
where,
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.08206atm.L/mol.K)
T is the absolute temperature (-14.0 °C + 273.15 = 259.2 K)
[tex]P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1atm . 41.0L}{(0.08206atm.L/mol.K).259.2K} =1.93mol[/tex]
According to the balanced equation, 6 moles of CO₂ are formed when 2 moles of C₃H₅N₃O₉ react. And the molar mass of nitroglycerin is 227 g/mol. Then, for 1.93 moles of CO₂:
[tex]1.93mol(CO_{2}).\frac{2mol(nitroglycerin)}{6mol(CO_{2})} .\frac{227g(nitroglycerin)}{1mol(nitroglycerin)} =146g(nitroglycerin)[/tex]
