Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t
ragic explosion at the Oklahoma City federal building in 1995. How many liters of gas at 307°C and 1.00 atm are formed by the explosive decomposition of 98.4 kg of ammonium nitrate to nitrogen, oxygen, and water vapor?
1 answer:
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.
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Answer:
215.12 g
Explanation:
The change in the bolling temperature is a collateral property called ebullioscopy. The elevation on temperature occurs because there'll be a solute in the pure compound, and that solute(generally non-volatile) has a different boiling point, so the total bolling pint should increase.
To calculate the elevation in the temperature, we must use the Raoult equation:
ΔT = KbxWxi
Where <em>ΔT</em> is the variation of the temperature of boiling point, <em>Kb</em> is the
ebullioscopy constant, <em>W</em> is the molarity, and <em>i</em> the Vant' Hoff factor, which for molecules is also equal to 1, because they don't form ions. So:
82.00 - 78.40 = 1.22xW
W = 3.60/1.22
W = 2.9508
The molarity can also be calculated by:
Where <em>m1</em> is the mass of the solute (in grames), <em>M1</em> is the molar mass of the solute, and <em>m2</em> is the mass of the solvent (in kg).
The density of ethanol is 0.789 g/cm³, and 1 mL = 1 cm³, so 600.00 mL = 600.00 cm³ :
0.789 = m2/600
m2 = 473.4 g = 0.4734 kg
The molar mass of geraniol is:
C: 12 g/mol x 10 = 120 g/mol
H: 1 g/mol x 18 = 18 g/mol
O: 16 g/mol x 1 = 16 g/mol
M1 = 154 g/mol
Then:
m1 = 215.12 g of geraniol.
1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.
2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.
3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.
4. 135 ml of stock HBr will be required to dilute the solution.
5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution
6. The pH of the resulting solution is 13.89
Explanation:
The formula used in solving the problems is
number of moles= 1st equation
molarity = 2nd equation
Dilution formula
M1V1 = M2V2 3rd equation
1. Data given
mass of Lithium nitrate = 21.6 grams
atomic mass of on emole lithium nitrate = 68.946 gram/mole
Molarity is given as 1.3 M
VOLUME=?
Calculate the number of moles using equation 1
n =
= 0.313 moles of lithium nitrate.
volume is calculated by applying equation 2.
volume =
= 0.240 litres of water will be used.
2. Data given:
mass of HCl = 215.1 gram
atomic mass of HCl = 36.46 gram/mole
volume = 2 litres
molarity = ?
using equation 1 number of moles calculated
number of moles =
number of moles of HCl = 5.899 moles
molarity is calculated by using equation 2
M =
= 2.9 M is the molarity of the solution of 2 litre HCl.
3. data given:
molarity of H2SO4 = 18 M
Solution to be made 250 ml of 6 M
USING EQUATION 3
18 x V1= 250 x 6
V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.
4. data given:
M1= 10M, V1 =?, M2= 3 ,V2= 450 ml
applying the equation 3
10 x VI = 3x 450
V1 = 135 ml of stock HBr will be required.
5. Data given:
V1 = 50 ml
M1= 12 M
V2=?
M2= 4
applying the equation 3
50 x 12 = 4 x v2
V2 = 150 ml.
6. data given:
HCl + NaOH ⇒ NaCl + H20
molarity of NaOH = 0.525 M
volume of NaOH = 25 ml
molarity of acid HCl= 75 ml
volume of HCl = 0.335 ml
pH=?
Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres
moles of NaOH = 0.0131
moles of HCl= 0.025 moles
The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.
difference of moles = 0.0119 moles ( NaOH moles is more)
Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution
molarity =
= 0.11 M (pOH Concentration)
14 = pH + pOH
pH = 14 - 0.11
pH = 13.89