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Nitroglycerin is an explosive used by the mining industry. It detonates according to the following balanced equation: 4C₃H₅N₃O₉(l) → 12CO₂(g) + 6N₂(g) + 10H₂O(g) + O₂(g) What volume is occupied by the carbon dioxide gas produced when 12.0 g of nitroglycerine explodes? The pressure is 1.25 atm at 523°C A=0.37 L B=8.29 L C=2.76 L D=1.52 L

a) 0.37 L
b) 8.29 L
c) 2.76 L
d) 1.52 L

1 Answer

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Final answer:

To find the volume of carbon dioxide produced from 12.0 g of nitroglycerin, calculate the moles of nitroglycerin, use the mole ratio from the balanced equation to find moles of CO₂, and then apply the ideal gas law with the given pressure and temperature.

Step-by-step explanation:

The volume of carbon dioxide gas produced by the detonation of nitroglycerin can be calculated using the ideal gas law, which is PV = nRT. Given that 12.0 g of nitroglycerin are used, we first need to calculate the number of moles of nitroglycerin, then use the stoichiometry of the balanced chemical reaction to find the moles of carbon dioxide produced. We then apply the ideal gas law using the given temperature and pressure, converting the temperature to Kelvin and using the appropriate gas constant (R).

First, we need the molar mass of nitroglycerin (molecular formula C₃H₅N₃O₉) which we use to convert grams to moles. Then, we use the balanced equation 4C₃H₅N₃O₉(l) → 12CO₂(g) + 6N₂(g) + 10H₂O(g) + O₂(g) to find the mole ratio between nitroglycerin and carbon dioxide, which is 4:12 or 1:3. After calculating the number of moles of CO₂, we can find the volume using the ideal gas law with the given pressure (1.25 atm) and temperature (796 K, which is 523°C + 273°C). Using R = 0.0821 L·atm/mol·K, we can solve for the volume (V).

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