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Answer:
The explosive decomposition of 98.4 kg of ammonium nitrate produces 58498.8 L of nitrogen, 29249.4 L of oxygen and 116997.6 L of water vapor.
Explanation:
To find how many liters of gas are formed from the explosive decomposition of ammonium nitrate it is necessary to follow these steps (check the attachment for better understanding):
1st) Balance the equation:
Write the decomposition equation and then find the correct coefficients to make sure that it goes according to the "Law of conservation of mass" (the mass of the reactants side must be equal to the mass of the products side). So, 2 moles of ammonium nitrate produces 2 moles of nitrogen, 1 mole of oxygen and 4 moles of water vapor.
2nd) Find the Ammonium nitrate molar mass:
The ammonium nitrate mass it is calculated by adding de molar mass of each atom that forms the ammonium nitrate molecule. You can find the elements molar mass in the Periodic Table.
In this example I decided to round the number to simplify tha calculus, for example: the oxygen molar mass in the periodic table is 15.9994 but I use 16. You can use the complete number if you want.
By doing this, the ammonium nitrate molar mass is 80 g/mol.
The statement says that there is 98.4 kg of ammonium nitrate. In ordder to use the same units in all the calculus sometimes it is usefull to convert the kg to g, so it is the same as 98400g. You can do it the other way around if you prefer (g to kg).
3rd) Find the number of moles of each gases and aqua vapor formed:
It is important to know the amount of each compound formed by the decomposition reaction, that's why we need to pay attention to the coefficients of the balanced reaction.
The amount of each compound is easily found by using the "rule of three".
To use the rule of three we need to think using the balanced reaction so:
If 160g (2 moles) of ammonium nitrate produces 2 moles of nitrogen gas, the 98400g that we have of ammonium nitrate will produce an X amount of nitrogen gas. With this information we multiply 98400g by 2 moles and then we divide the result by 160g. The final result it is 1230 moles of nitrogen.
In the same way we use the rule of three to calculate the number of moles of oxygen and water.
4th) Find the liters (volume) of each gas and aqua vapor formed:
Finally, to find the liters from the number of moles, it is necessary to apply the "Ideal gases law", that relates the pressure (atm), volume (L), moles number and temperature (Kelvin) with the R gas constant in the formula:
PxV = nxRxT
It is important to use the correct units because the R gas constant is equal to 0.082 atm.L/mol.K.
As we need to calculate the liters (volume) we pass the pressure dividing to the other side and then we just have to replace the information:
V = (nxRxT)/P
As you can see in the attachment, doing this last step for each compound, we can find the liters produced of them.
Considering the definition of molar mass, the moles of gas used are 10.625 moles.
<h3>Definition of molar mass</h3>The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
<h3>Amount of moles used</h3>Natural gas has a molar mass of 16.0 g/mole.
You started out the day with a tank containing 200.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 200 grams are contained in how many moles?
<u><em>amount of moles at the beginning= 12.5 moles</em></u>
At the end of the day, your tank contains 30.0 g of natural gas. So, you can apply the following rule of three: If by definition of molar mass 16 grams are contained in 1 mole, 30 grams are contained in how many moles?
<u><em>amount of moles at the end= 1.875 moles</em></u>
The number of moles used will be the difference between the number of moles used initially and the contents at the end of the day.
moles used= amount of moles at the beginning - amount of moles at the end
moles used= 12.5 moles - 1.875 moles
<u><em>moles used= 10.625 moles</em></u>
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Finally, the moles of gas used are 10.625 moles.
Learn more about molar mass:
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