Answer:
0.0177 L of nitrogen will be produced
Explanation:
The decomposition reaction of sodium azide will be:
![2NaN_{3}(s)--->2Na(s)+3N_{2}(g)](https://tex.z-dn.net/?f=2NaN_%7B3%7D%28s%29---%3E2Na%28s%29%2B3N_%7B2%7D%28g%29)
As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas
The molecular weight of sodium azide = 65 g/mol
The mass of sodium azide used = 100 g
The moles of sodium azide used = ![\frac{mass}{molarmass}=\frac{100}{65}=1.54mol](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bmolarmass%7D%3D%5Cfrac%7B100%7D%7B65%7D%3D1.54mol)
so 1.54 moles of sodium azide will give =
mol
the volume will be calculated using ideal gas equation
PV=nRT
Where
P = Pressure = 1.00 atm
V = ?
n = moles = 2.31 mol
R = 0.0821 L atm / mol K
T = 25 °C = 298.15 K
Volume = ![\frac{P}{nRT}=\frac{1}{2.31X0.0821X298.15}=0.0177L](https://tex.z-dn.net/?f=%5Cfrac%7BP%7D%7BnRT%7D%3D%5Cfrac%7B1%7D%7B2.31X0.0821X298.15%7D%3D0.0177L)
Answer:
This is a coal combustion process and we will assume
Inlet coal amount = 100kg
It means that there are
15kg of H2O, 2kg of Sulphur and 83kg of Carbon
Now to find the mole fraction of SO2(g) in the exhaust?
Molar mass of S = 32kg/kmol
Initial moles n of S = 2/32 = 0.0625kmols
Reaction: S + O₂ = SO₂
That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂
Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust
The mole fraction of SO2(g) in the exhaust=0.0625kmols
Explanation:
I would always start by balancing your carbons, and then balancing the rest from there.
1. C2H5OH + O2 —> CO2 + H2O - You have two carbons on the left and one on the right. Multiply CO2 by 2.
C2H5OH + O2 —> 2CO2 + H2O
Now balance hydrogen. You have 6 on the left and 2 on the right. Multiply H2O by 3.
C2H5OH + O2 —> 2CO2 + 3H2O
Now balance oxygen. You have 3 on the left and 7 on the right. You need 4 more on the left. Don’t multiply the C2H5OH by anything because that will change the numbers of everything else too. Multiply O2 by 3 instead.
C2H5OH + 3O2 —> 2CO2 + 3H2O
Check that all atoms are now balanced, and you’re good.
2. Same process as before.
First carbons - C3H8 + O2 —> 3CO2 + H2O
Then hydrogens - C3H8 + O2 —> 3CO2 + 4H2O
Then oxygens - C3H8 + 5O2 —> 3CO2 + 4H2O
3. Same again.
Carbons) C6H12O6 + O2 —> 6CO2 + H2O
Hydrogens) C6H12O6 + O2 —> 6CO2 + 6H2O
Oxygens) C6H12O6 + 6O2 —> 6CO2 + 6H2O
4. The general reaction for a combustion reaction is a hydrocarbon reacting with oxygen to produce carbon dioxide and water.
A "3" should but put in front of
<span>"cas o 4 "</span>