Answer:
742.2 L
Explanation:
First we must find the number of moles of nitroglycerine reacted.
Molar mass of nitroglycerine= 227.0865 g/mol
Mass of nitroglycerine involved = 1×10^3 g
Number of moles of nitroglycerine= 1×10^3g/227.0865 g/mol
n= 4.40361 moles
T= 1985°C + 273= 2258K
P= 1.100atm
R= 0.082atmLmol-1K-1
Using the ideal gas equation:
PV= nRT
V= nRT/P
V= 4.40361× 0.082× 2258/1.1
V= 742 L
Considering the reaction stoichiometry and ideal gas law, the volume of gas produced is 5375.626 L.
The balanced reaction is:
4 C₃H₅N₃O₉(s) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Then 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas [12 moles of CO₂(g) + 10 moles of H₂O(g) + 6 moles of N₂(g) + 1 mole O₂(g)]
Being the molar mass of nitroglycerine C₃H₅N₃O₉(s) 227 g/mole, then the amount of moles that 1 kg (1000 g) of the compound contains can be calculated as:
[tex]1000 gramsx\frac{1 mole}{227 grams}= 4.405 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas, 4.405 moles of C₃H₅N₃O₉(s) will produce how many moles of gas?
[tex]amount of moles of gas=\frac{4.405 moles of nitroglycerinex29 moles of gas}{4 moles of nitroglycerine}[/tex]
amount of moles of gas= 31.93625 moles
On the other hand, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
Replacing:
1.1 atm× V= 31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K
Solving:
V= (31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K) ÷ 1.1 atm
V= 5375.625 L
Finally, the volume of gas produced is 5375.626 L.
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